proof that diagonals of a quadrilateral are perpendicular if $AB^2+CD^2=BC^2+AD^2$ proof that diagonals of a quadrilateral are perpendicular if $AB^2+CD^2=BC^2+AD^2$.
My Attempt:we know that if diagonals of a quadrilateral are perpendicular then we have $AB^2+CD^2=BC^2+AD^2$.But have to proof opposite of it?


Assume the intersection of the to diagonals is $O$. Let $|OA|=a,|OB|=b,|OC|=c,|OD|=d$. Assume $\angle AOB=\gamma$. Then 
$$|AB|^2=a^2+b^2-2ab \cos\gamma,$$
$$|CD|^2=c^2+d^2-2cd \cos\gamma,$$
$$|BC|^2=b^2+c^2-2bc \cos(\pi-\gamma)=b^2+c^2+2bc \cos\gamma,$$
$$|AB|^2=a^2+d^2-2ad \cos(\pi-\gamma)=a^2+d^2+2ad \cos\gamma.$$
From the condition, we have $-2ab \cos\gamma-2cd \cos\gamma=2bc \cos\gamma+2ad \cos\gamma$. Thus, $(a+c)(b+d)\cos\gamma=0\Rightarrow \cos\gamma=0\Rightarrow \gamma=\frac{\pi}{2}$. Thus, $AC\perp BD$.
