# Given #ABC# a triangle where #bar(AD)# is the median and let the segment line #bar(BE)# which meets #bar(AD)# at #F# and #bar(AC)# at #E#. If we assume that #bar(AE)=bar(EF)#, show that #bar(AC)=bar(BF)#?.

See below.

We will apply the theorem of Menelaus of Alexandria to the sub-triangle

According to Menelaus,

but

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Since AD is a median,Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

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Using theGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

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Using the propertyGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC intoGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property ofGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

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Using the property of medGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equalGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of mediansGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segmentsGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians inGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments,Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in aGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, sayGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangleGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BDGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle,Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD andGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we knowGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DCGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD dividesGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC.Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC intoGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. ThusGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equalGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus,Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segmentsGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, weGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments.Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we haveGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. LetGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let'sGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

1Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote theGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

1.Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote the lengthGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BDGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote the length ofGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD =Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote the length of ADGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DCGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote the length of AD asGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote the length of AD as mGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

NowGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote the length of AD as m.Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now,Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote the length of AD as m. ThereforeGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AEGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote the length of AD as m. Therefore,Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote the length of AD as m. Therefore, BDGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF,Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote the length of AD as m. Therefore, BD =Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can considerGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote the length of AD as m. Therefore, BD = DCGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote the length of AD as m. Therefore, BD = DC =Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF andGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote the length of AD as m. Therefore, BD = DC = mGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote the length of AD as m. Therefore, BD = DC = m/Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote the length of AD as m. Therefore, BD = DC = m/2Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote the length of AD as m. Therefore, BD = DC = m/2.

Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

2Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote the length of AD as m. Therefore, BD = DC = m/2.

SinceGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

2.Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Using the property of medians in a triangle, we know that AD divides BC into two equal segments. Let's denote the length of AD as m. Therefore, BD = DC = m/2.

Since AEGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AEGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE =Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE =Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF =Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, asGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF =Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is theGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the otherGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of ADGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

NowGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangleGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (VertGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (VerticallyGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF.Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically oppositeGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. SinceGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite anglesGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles) Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF =Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles) 4Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FEGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles) 4.Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AEGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE =Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF,Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF =Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF isGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF is anGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF is an isGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF is an isosGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (CorGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF is an isosceGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (CorrespondGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF is an isoscelesGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (CorrespondingGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF is an isosceles triangleGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding anglesGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF is an isosceles triangle.Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF is an isosceles triangle. ThereforeGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

TrianglesGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF is an isosceles triangle. Therefore,Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF is an isosceles triangle. Therefore, theGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF is an isosceles triangle. Therefore, the anglesGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF andGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF is an isosceles triangle. Therefore, the angles atGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF is an isosceles triangle. Therefore, the angles at A andGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF is an isosceles triangle. Therefore, the angles at A and EGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congrGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF is an isosceles triangle. Therefore, the angles at A and E areGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruentGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF is an isosceles triangle. Therefore, the angles at A and E are equal.

Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent byGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF is an isosceles triangle. Therefore, the angles at A and E are equal.

SinceGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by theGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Now, consider triangle AEF. Since AF = FE and AE = EF, triangle AEF is an isosceles triangle. Therefore, the angles at A and E are equal.

Since ADGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the SideGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is aGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-AGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a medianGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-AngleGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median,Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-SideGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it dividesGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BCGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC intoGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SASGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into twoGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS)Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equalGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterionGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segmentsGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion.Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments.Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. ThereforeGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. HenceGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence,Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

5Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angleGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

5.Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BADGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts ofGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD =Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congrGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angleGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruentGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CADGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent trianglesGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

SinceGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now,Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since ADGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, considerGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD isGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangleGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is aGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a medianGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median,Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF andGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median, itGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median, it divides BCGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. TheyGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median, it divides BC intoGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They shareGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median, it divides BC into twoGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share theGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median, it divides BC into two equalGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the sideGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median, it divides BC into two equal segmentsGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median, it divides BC into two equal segments.Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF andGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median, it divides BC into two equal segments. ThisGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and alsoGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median, it divides BC into two equal segments. This impliesGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also theGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median, it divides BC into two equal segments. This implies thatGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angleGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median, it divides BC into two equal segments. This implies that ACGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle atGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median, it divides BC into two equal segments. This implies that AC andGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at AGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median, it divides BC into two equal segments. This implies that AC and BFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median, it divides BC into two equal segments. This implies that AC and BF areGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median, it divides BC into two equal segments. This implies that AC and BF are alsoGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

UsingGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median, it divides BC into two equal segments. This implies that AC and BF are also equal.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using theGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median, it divides BC into two equal segments. This implies that AC and BF are also equal.

6Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the AngleGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

Since AD is a median, it divides BC into two equal segments. This implies that AC and BF are also equal.

6.Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-SGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- ACGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-SideGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC =Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-Side-AGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-Side-Angle (Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (FromGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-Side-Angle (ASA)Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-Side-Angle (ASA) similarity criterionGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 andGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-Side-Angle (ASA) similarity criterion,Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the propertyGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-Side-Angle (ASA) similarity criterion, weGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property ofGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-Side-Angle (ASA) similarity criterion, we canGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-Side-Angle (ASA) similarity criterion, we can conclude thatGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of mediansGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-Side-Angle (ASA) similarity criterion, we can conclude that triangleGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medians)

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-Side-Angle (ASA) similarity criterion, we can conclude that triangle BGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medians)

HGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-Side-Angle (ASA) similarity criterion, we can conclude that triangle BAFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medians)

HenceGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-Side-Angle (ASA) similarity criterion, we can conclude that triangle BAF isGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medians)

Hence,Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-Side-Angle (ASA) similarity criterion, we can conclude that triangle BAF is similarGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medians)

Hence, weGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-Side-Angle (ASA) similarity criterion, we can conclude that triangle BAF is similar toGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medians)

Hence, we'veGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-Side-Angle (ASA) similarity criterion, we can conclude that triangle BAF is similar to triangleGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medians)

Hence, we've shownGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-Side-Angle (ASA) similarity criterion, we can conclude that triangle BAF is similar to triangle CAF.

As aGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medians)

Hence, we've shown that if AE = EFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-Side-Angle (ASA) similarity criterion, we can conclude that triangle BAF is similar to triangle CAF.

As a result ofGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medians)

Hence, we've shown that if AE = EF,Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

Using the Angle-Side-Angle (ASA) similarity criterion, we can conclude that triangle BAF is similar to triangle CAF.

As a result of theGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medians)

Hence, we've shown that if AE = EF, thenGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

As a result of the similarityGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medians)

Hence, we've shown that if AE = EF, then ACGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

As a result of the similarity,Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medians)

Hence, we've shown that if AE = EF, then AC =Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

As a result of the similarity, theGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medians)

Hence, we've shown that if AE = EF, then AC = BFGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

As a result of the similarity, the corresponding sidesGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medians)

Hence, we've shown that if AE = EF, then AC = BF inGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

As a result of the similarity, the corresponding sides areGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medians)

Hence, we've shown that if AE = EF, then AC = BF in triangleGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

As a result of the similarity, the corresponding sides are inGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medians)

Hence, we've shown that if AE = EF, then AC = BF in triangle ABCGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

As a result of the similarity, the corresponding sides are in proportion.Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medians)

Hence, we've shown that if AE = EF, then AC = BF in triangle ABC.Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

As a result of the similarity, the corresponding sides are in proportion. ThusGiven triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF, we need to prove that AC = BF.

Since AD is a median, it divides BC into two equal segments, say BD and DC. Thus, we have:

- BD = DC

Now, because AE = EF, we can consider triangles AEF and BEF:

- AE = EF (Given)
- ∠AEF = ∠BEF (Vertically opposite angles)
- ∠EAF = ∠EBF (Corresponding angles)

Triangles AEF and BEF are congruent by the Side-Angle-Side (SAS) criterion. Therefore:

- AF = BF (Corresponding parts of congruent triangles)

- AC = BF (From 5 and the property of medians)

Hence, we've shown that if AE = EF, then AC = BF in triangle ABC.Given triangle ABC with median AD, and segment BE intersecting AD at F and AC at E such that AE = EF. We want to show that AC = BF.

Since AE = EF, this implies that AF = FE, as AF is the other half of AD.

Since AD is a median, it divides BC into two equal segments. Hence, angle BAD = angle CAD.

Now, consider triangle BAF and triangle CAF. They share the side AF and also the angle at A.

As a result of the similarity, the corresponding sides are in proportion. Thus, we have:

[ \frac{AC}{BF} = \frac{CA}{AF} = \frac{AC}{AF} ]

Given that AF = FE, we can write:

[ \frac{AC}{BF} = \frac{AC}{AF + FE} = \frac{AC}{AD} ]

Since AD is the median of triangle ABC and BD = DC = m/2:

[ AD = 2 \times \frac{m}{2} = m ]

Therefore, ( \frac{AC}{BF} = \frac{AC}{AD} = 1 )

This implies that AC = BF. Hence, we have shown that if AE = EF, then AC = BF.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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