D is a point on the side BC of a triangle ABC such that ∠ ADC = ∠ BAC , prove that CA2 = CB.CD

Sol.
In $\Delta ABC$, D is a point on side BC such that $\angle ADC = \angle BAC$
In $\Delta CBA$ and $\Delta CDA$
$\angle C = \angle C$ (common)
$\angle BAC = \angle ADC$ (given)
$\therefore \Delta CBA \sim \Delta CAD$ (By AA similarity)
$\therefore$ their corresponding sides are proportional
$\frac{CB}{CA} = \frac{CA}{CD} \Rightarrow CA^2 = CB. CD$