# Let #hat(ABC)# be any triangle, stretch #bar(AC)# to D such that #bar(CD)≅bar(CB)#; stretch also #bar(CB)# into E such that bar(CE)≅bar(CA). Segments #bar(DE) and bar(AB)# meet at F. Show that #hat(DFB# is isosceles?

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Given any triangle #hat(ABC)# , stretch #bar(AC)# to D such that #bar(CD)≅bar(CB)# ; stretch also #bar(CB)# into E such that bar(CE)≅bar(CA).Segment #bar(DE) and bar(AB)# meet at F. Show that #hat(DFB# is isoscheles?

Given any triangle

As follows

Ref:Given Figure

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To show that triangle DFB is isosceles, we'll prove that segment DF is congruent to segment FB.

Since CE is stretched to be congruent to CA, we have CE ≅ CA. And since CD is stretched to be congruent to CB, we have CD ≅ CB.

In triangle ABC, by the Segment Addition Postulate: AC = AE + EC AB = AD + DC

Since EC = CA and DC = CB: AC = AE + CA AB = AD + CB

Rearranging terms: AC - CA = AE AB - CB = AD

But AC - CA = AE is the same as AE = CE, and AB - CB = AD is the same as AD = DB.

Now, in triangle DEF: Using the triangle proportionality theorem (also known as the Thales' Theorem), we can state that: AD/DB = AE/EC

Substitute the values we derived earlier: DB/DB = AE/EC

This simplifies to: 1 = 1

Since the ratios are equal, we have: AD = DB

This means that DF = FB, thus proving that triangle DFB is isosceles.

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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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