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?

 A: 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, $AC\perp BD$.
A: 
Lemma. Let $ABC$ be a triangle and $H_A$ be the projection of $A$ on $BC$. For any $P\in AH_A$, we have:  $$PB^2-PC^2 = AB^2 - AC^2 = H_A B^2-H_A C^2$$
   by the  Pythagorean theorem.

If $AC\perp BD$, the projections of $A$ and $C$ on $BD$ are the same point, hence:
$$ AB^2-AD^2 = CB^2-CD^2 $$
and the claim follows. In the opposite way: if $AB^2-AD^2=CB^2-CD^2$, the projections of $A$ and $C$ on $BD$ are the same point, hence $AC\perp BD$.
A: $$AB^2+CD^2=BC^2+AD^2 \Leftrightarrow AB^2-AD^2=CB^2-CD^2 \Leftrightarrow$$
$$\Leftrightarrow \vec{AB}^2-\vec{AD}^2=\vec{CB}^2-\vec{CD}^2\Leftrightarrow$$
$$\Leftrightarrow \left(\vec{AB}-\vec{AD}\right)\left(\vec{AB}+\vec{AD}\right)=\left(\vec{CB}-\vec{CD}\right)\left(\vec{CB}+\vec{CD}\right)\Leftrightarrow$$
$$\Leftrightarrow \left(\vec{AB}+\vec{AD}\right)\vec{DB}=\left(\vec{CB}+\vec{CD}\right)\vec{DB}\Leftrightarrow$$
$$\Leftrightarrow \vec{DB}\left(\vec{AB}+\vec{AD}-\vec{CB}-\vec{CD}\right)=0\Leftrightarrow$$
$$\Leftrightarrow 2\vec{AC}\cdot\vec{DB}=0\Leftrightarrow AC \perp BD$$
