In a quadrilateral ABCD, $∠ B = 90^{\circ}, A D^{2} = A B^{2} + B C^{2} + C D^{2}$ , prove that $∠ A C D = 90^{\circ}$

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Solution

From Pythagoras theorem, $A C^{2} = A B^{2} + B C^{2}$

Given that $A D^{2} = A B^{2} + B C^{2} + C D^{2}$

$\Rightarrow A D^{2} = A C^{2} + C D^{2}$

Now, from converse of Pythagoras theorem, in $△ A C D$ , angle opposite to AD is $90^{\circ}$ .

$∴ ∠ A C D = 90^{\circ}$

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A B C D

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∠ B = 90^{\circ} a n d A D^{2} = A B^{2} + B C^{2} + C D^{2} . t h e n ∠ A C D i s

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