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