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

Given:

In a quadrilateral \( A B C D, \angle B=90^{\circ}, A D^{2}=A B^{2}+B C^{2}+C D^{2} \).

To do:

We have to prove that $\angle A C D=90^o$.

Solution:

In $\triangle ABC$, by Pythagoras theorem,

$AC^2=AB^2+BC^2$.....(i)

$AD^{2}=AB^{2}+BC^{2}+CD^{2}$

$AD^2=AC^2+CD^2$    (From (i))

This implies, by Converse of Pythagoras theorem,

$\triangle ACD$ is a right-angled triangle with right angle at $C$.

Therefore, $\angle ACD=90^o$

Hence proved.

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Updated on: 10-Oct-2022

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