Question:
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 $\angle A C D=90^{\circ}$.
Solution:
In order to prove angle $\angle A C D=90^{\circ}$ it is enough to prove that $A D^{2}=A C^{2}+C D^{2}$.
Given, $A D^{2}=A B^{2}+B C^{2}+C D^{2}$
$A D^{2}-C D^{2}=A B^{2}+B C^{2}$.....(1)
Since $\angle B=90^{\circ}$, so applying Pythagoras theorem in the right angled triangle $\mathrm{ABC}$, we get
$A C^{2}=A B^{2}+B C^{2}$......(2)
From (1) and (2), we get
$A C^{2}=A D^{2}-C D^{2}$
$A C^{2}+C D^{2}=A D^{2}$
Therefore, angle $\angle A C D=90^{\circ}$ (Converse of pythagoras theorem)