Question
In a quadrilateral $ABCD$, $\angle\text{B}=90^\circ,$ $AD^2 = AB^2 + BC^2 + CD^2$, prove that $\angle\text{ACD}=90^\circ.$
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Answer
In order to prove angle $\angle\text{ACD}=90^\circ$ it is enough to prove that $AD^2 = AC^2 + CD^2$
Given: $AD^2 = AB^2 + BC^2 + CD^2$
$AD^2 - CD^2 = AB^2 + BC^2 ......(1)$
Since $\angle\text{B}=90^\circ,$ so applying pythagoras theorem in the right angled triangle $ABC$, we get,
$AC^2 = AB^2 + BC^2 ....(2)$
From (1) and (2), we get
$AC^2 = AD^2 - CD^2$
$AC^2 + CD^2 = AD^2$
Therefore, angle $\triangle\text{ACD}=90^\circ.$ (Converse of pythagoras theorem)
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