Question

In triangle ABC ,AB = AC . D is a point of ac such that BC^2 = ACxCD . Prove BD = BC

Answer 1

ΔABC, AB = AC D is a point on AC such that BC2 = AC × AD  In ΔABC and ΔBDC  ∠C = ∠C (Common angle) ∴ ΔABC ~ ΔBDC [By SAS similarity criterion]  [Since triangles are similar, corresponding sides are proportional]  From (1) and (2), we get  ∴ BC = BD
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