In a quadrilateral ABCD ∠B=90∘AD2=AB2+BC2+CD2 prove tha

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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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 $90^{\circ}$ .

$∴ ∠ A C D = 90^{\circ}$