In $△ A B C$ , $∠ A = 90^{0}, C A = A B$ and $D$ is a point on $A B$ produced.
Prove that $D C^{2} - B D^{2} = 2 A B . A D$

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Solution

In $Δ C A D,$
$C D^{2} = {A C}^{2} + {A D}^{2}$
$C D^{2} = {A B}^{2} + {A D}^{2}$ --(1) ....( $∵ A B = A C$ )

$A D = A B + B D B D = A D - A B {B D}^{2} = {A D}^{2} + {A B}^{2} - 2 A B . A D$ --(2)

Substituting (1) in (2),
${B D}^{2} = C D^{2} - 2 A B . A D C D^{2} - {B D}^{2} = 2 A B . A D$

Hence proved.

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A B C

A = 90^{o}, C A = A B

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In a triangle ABC, angle A=90°, CA=AB and D is a point on AB produced. Prove that

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