# In a triangle ABC, AB=AC and D is a point on side AC such that BCxBC=ACxCD. How to prove that BD=BC?

Given

#"In "DeltaABC#

#AB=AC and D" is a point on"AC " such that"#

#BCxxBC=ACxxAD#

#"We are to prove "BD=BC# Proof

Rearrenging the given relation

#BCxxBC=ACxxAD" "# We can write

#(BC)/(CD)=(AC)/(BC)->DeltaABC" similar "DeltaBDC#

Their corresponding angle pairs are:

#1. /_BAC "= corresponding "/_DBC#

#2. /_ABC "= corresponding "/_BDC#

#3. /_ACB " =corresponding "/_DCB#

So as per above relation 2 we have

#/_ABC =" corresponding "/_BDC#

#"Again in"DeltaABC#

#AB=AC->/_ABC=/_ACB=/_DCB#

#:."In "DeltaBDC,/_BDC=/_BCD#

#->BD=BC#

Alternative way

The ratio of corresponding sides may be written in extended way as follows

#(BC)/(CD)=(AC)/(BC)=(AB)/(BD)#

From this relation we have

#(AC)/(BC)=(AB)/(BD)#

#=>(AC)/(BC)=(AC)/(BD)->"As "AB=AC" given"#

#=>1/(BC)=1/(BD)#

#=>BC=BD#

ProvedHope, this will help

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*Answer 2Sign up to view the whole answerSign up with email*Given triangle (ABC) with (AB = AC) and a point (D) on side (AC) such that (BC \cdot BC = AC \cdot CD), we want to prove that (BD = BC).

Since (AB = AC), triangle (ABC) is isosceles. Let (E) be the midpoint of (BC).

Now, from the given condition, (BC \cdot BC = AC \cdot CD), implying that (BC^2 = AC \cdot CD). Since (AB = AC), we can rewrite this as (BC^2 = AB \cdot CD).

Using the Power of a Point Theorem, we can write (BD \cdot CD = AD \cdot DC).

Since triangle (ABC) is isosceles, (AE) is perpendicular to (BC), so (AE) is also the median, altitude, and angle bisector of (ABC). Thus, triangles (ABE) and (ACE) are congruent by the hypotenuse-leg congruence criterion. Hence, (BE = EC).

Now, consider triangle (BDE). It is a right triangle with (BE = EC), so (BD = DE).

Considering triangle (CDE), we have (CD = CD), and since (BD = DE), by the Hypotenuse-Leg congruence criterion, triangles (BDE) and (CDE) are congruent. This implies (BC = CE).

Now, since (BC = CE) and (BE = EC), we have (BC = BE + EC = BD).

Hence, (BD = BC).

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*Answer from HIX Tutor**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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