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?

Given: A △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 △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. 
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Hope this helps....
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