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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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