Question:
In a quadrilateral $\mathrm{ABCD}, \angle \mathrm{B}=90^{\circ}$. If $\mathrm{AD}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CD}^{2}$ then prove that $\angle \mathrm{ACD}=90^{\circ}$.
Solution:
In quadrilateral $A B C D$, we have
$\angle B=90^{\circ}$
So, $A C^{2}=A B^{2}+B C^{2} \quad$ (Pythagoras theorem)
and
$A D^{2}=A B^{2}+B C^{2}+C D^{2} \quad$ (Given)
So,
$A D^{2}=A B^{2}+B C^{2}+C D^{2}$
$A D^{2}=A C^{2}+C D^{2}$
Hence, $\angle A C D=90^{\circ}$ (Converse of Pythagoras theorem)