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.

Given: A Δ ABC in which AB = AC. D is a point on AC such that BC² = AC × CD. 
                                      
To prove: BD = BC
Proof: Since BC² = AC × CD
Therefore BC × BC = AC × CD
AC/BC = BC/CD ……………(i)
Also ∠ACB = ∠BCD
Since ΔABC ∼ ΔBDC [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.