Question

In triangle ABC point D is on side BC (B-D-C) such that ∠BAC = ∠ADC then prove that CA² = CB × CD

Answer 1

Proof:

In ΔBAC and ΔADC,

∠BAC ≅ ∠ADC   ...[Given]

∠BCA ≅ ∠ACD   ...[Common angle]

∴ ΔBAC ∼ ΔADC   ...[AA test of similarity]

∴ `(CA)/(CD) = (CB)/(CA)`   ...[Corresponding sides of similar triangles]

∴ CA × CA = CB × CD

∴ CA² = CB × CD