ABC is an isosceles triangle with ab is equal to AC and D is a point On AC such that BC square is equal to AC into CD prove that BD is equal to BC
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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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