Question

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

Answer 1

$\begin{array}{c}InthequadrilateralABCD\\ \angle ABC={90}^{\circ }\\ Hence,A{B}^2+B{C}^2=A{C}^2----\left(1\right)Byu\sin gpythagorastheroem\\ Now,\\ A{D}^2=\left(A{B}^2+B{C}^2\right)+C{D}^2\left(given\right)\\ =A{C}^2+C{D}^2---\left(2\right)\\ InthetriangleACD,ADhastobeahypotenuseand\\ \angle ACD={90}^{\circ }bypythagorastheorem\\ proved\end{array}$