#
In a triangle ABC, AB=AC and D is point on AC such that BC^2=AC*CD.

Prove that BD=DC using Pythagoras theorem?

^{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.