**ABC is a triangle in which AB = Ac and D is a point on the side AC such that BC2 = AC × CD. Prove that BD = BC.**

###### Answers:

Given: A △ABC in which AB = AC. D is a point on AC such that BC^{2} = AC × CD.

To prove : BD = BC

Proof : Since BC^{2} = AC × CD

Therefore BC × BC = AC × CD

AC/BC = BC/CD …….(i)

Also ∠ACB = ∠BCD

Since △ABC ~ △BDC [By SAS Axiom of similar triangles]

AB/AC = BD/BC ……..(ii)

But AB = AC (Given) ………(iii)

From (i),(ii) and (iii) we get

BD = BC.