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

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Answer 1 · 2016-08-24T16:30:35

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

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

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

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

$=>BC=BD$

Proved

Hope, this will help