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:
![](/assets/questions/media/153848-59944-1646928321.png)
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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