$A B C D$ is a rectangle. ' $P$ ' is any point outside it such that ${P A}^{2} + {P C}^{2} = {B A}^{2} + {A D}^{2}$ . Prove that $∠ A P C = 90^{o}$ .

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Solution

Join $A C$

In $△ A B C$ ,

$A C^{2} = A B^{2} + B C^{2}$
$\Rightarrow A C^{2} = A B^{2} + A D^{2} . . . . . (1)$ (as $B C = A D$ (parallel sides))

It is given that $P A^{2} + P C^{2} = B A^{2} + A D^{2}$ , therefore, using equation 1, we have

$P A^{2} + P C^{2} = A C^{2}$

Since $P A^{2} + P C^{2} = A C^{2}$ , therefore $△ A P C$ is a right angled triangle.

Hence, $∠ A P C = 90^{0}$ .

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