Question

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

Answer 1

 Given in 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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