Triangle $A B C$ is right angled at $C$ and $C D$ is perpendicular to $A B$ . Prove that ${B C}^{2} \times A D = {A C}^{2} \times B D$ .

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Solution

Consider $△ A C D & △ A B C$
$∠ C A D = ∠ C A B$
$∠ C D A = ∠ A C B$
$∴ △ A C D \sim △ A B C,$ { By AA similarty criterion}
$⟹ \frac{A C}{A B} = \frac{A D}{A C}$
$∴ {A C}^{2} = A B \times A D - (1)$
Similarily, $△ B C D \sim △ B A C$ { by AA similarity criterion}
$⟹ \frac{B C}{B A} = \frac{B D}{B C}$
$∴ {B C}^{2} = B A \times B D - (2)$
$∴ \frac{{B C}^{2}}{{A C}^{2}} = \frac{A B \times B D}{A B \times A D}$
$∴ {B C}^{2} \times A D = {A C}^{2} \times B D$

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