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In the given figure, D is a point on the side BC of $\triangle ABC$ such that $\angle ADC = \angle BAC$. Show that $CA^2 = CD.CB$.
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In $\triangle ACD$ and $\triangle BCA$
$\angle ADC = \angle BAC$
$\angle ACD = \angle BCA$
$\therefore \triangle ACD \sim \triangle BCA$
So, $\frac{CA}{CB} = \frac{CD}{CA}$
$\Rightarrow CA^2 = CD.CB$
$\angle ADC = \angle BAC$
$\angle ACD = \angle BCA$
$\therefore \triangle ACD \sim \triangle BCA$
So, $\frac{CA}{CB} = \frac{CD}{CA}$
$\Rightarrow CA^2 = CD.CB$