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)