In a triangle ABC , AB=AC and D is a point on AC such that (BC)^2 = AC*CD . Prove that BD =BC
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Step-by-step explanation:
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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