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# D is a point on the side BC of a triangle ABC, such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB.CD$.

Given:

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

To do:

We have to show that $CA^2 = CB.CD$.

Solution:

In $\triangle ABC$ and $\triangle DAC$,

$\angle C=\angle C$ (common)

$\angle BAC=\angle ADC$

Therefore, by AA criterion,

$\triangle \mathrm{ABC} \sim \triangle \mathrm{DAC}$

This implies,

$\frac{\mathrm{CA}}{\mathrm{CD}}=\frac{\mathrm{CB}}{\mathrm{CA}}$

This implies,

$CA^2=CB \times CD$

Hence proved.

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