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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dk_ch Share
Aug 25, 2016

Drawn

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

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