Question

In the adjoining figure, in ∆ABC, point D is on side BC such that, ∠BAC = ∠ADC. Prove that, CA² = CB × CD,

Answer 1

Proof:

In ∆BAC and ∆ADC,

∠BAC ≅ ∠ADC [Given] 

∠BCA ≅ ∠ACD [Common angle] 

∴ ∆BAC ~ ∆ADC [AA test of similarity]

∴ \(\frac{CA}{CD} = \frac{CB}{CA}\) [Corresponding sides of similar triangles]

∴ CA × CA = CB × CD 

∴ CA²  = CB × CD