Question

D is a point on the side BC of a triangle ABC such that $\angle$ADC = $\angle$BAC. Show that CA2=CB.CD

Answer 1

Given: $\angle ADC = \angle BAC$

To Prove: $CA^{2}=CB.CD$

Proof: In ∆ADC and ∆BAC,
$\angle$ADC = $\angle$BAC (Given)
$\angle$ACD = $\angle$BCA (Common angle)
∴ ∆ADC ∼ ∆BAC (By AA similarity criterion)

We know that the corresponding sides of similar triangles are in proportion.
∴$\frac{CA}{CB}=\frac{CD}{CA}$
$⇒CA^2=CB \times CD$

Hence Proved