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