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$D$ is a point on side $BC$ of $\Delta ABC$ such that $\angle ADC = \angle BAC$. Prove that $(CA)^2 = CB \cdot CD$.
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Solution:
In $\Delta ADC$ and $\Delta BAC$,
$\angle ADC = \angle BAC$ (given)
$\angle C = \angle C$ (common) [1/2 mark]
$\Delta ADC \sim \Delta BAC$ (By AA similarity criterion) [1/2 mark]
$\Rightarrow \frac{CA}{CB} = \frac{CD}{CA} \Rightarrow (CA)^2 = CB \cdot CD$ [1/2 mark]
Correct figure: [1/2 mark]
In $\Delta ADC$ and $\Delta BAC$,
$\angle ADC = \angle BAC$ (given)
$\angle C = \angle C$ (common) [1/2 mark]
$\Delta ADC \sim \Delta BAC$ (By AA similarity criterion) [1/2 mark]
$\Rightarrow \frac{CA}{CB} = \frac{CD}{CA} \Rightarrow (CA)^2 = CB \cdot CD$ [1/2 mark]
Correct figure: [1/2 mark]