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?



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 $AC\perp BD$.
