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?


$$AB^2+CD^2=BC^2+AD^2 \Leftrightarrow AB^2-AD^2=CB^2-CD^2 \Leftrightarrow$$
$$\Leftrightarrow \vec{AB}^2-\vec{AD}^2=\vec{CB}^2-\vec{CD}^2\Leftrightarrow$$
$$\Leftrightarrow \left(\vec{AB}-\vec{AD}\right)\left(\vec{AB}+\vec{AD}\right)=\left(\vec{CB}-\vec{CD}\right)\left(\vec{CB}+\vec{CD}\right)\Leftrightarrow$$
$$\Leftrightarrow \left(\vec{AB}+\vec{AD}\right)\vec{DB}=\left(\vec{CB}+\vec{CD}\right)\vec{DB}\Leftrightarrow$$
$$\Leftrightarrow \vec{DB}\left(\vec{AB}+\vec{AD}-\vec{CB}-\vec{CD}\right)=0\Leftrightarrow$$
$$\Leftrightarrow 2\vec{AC}\cdot\vec{DB}=0\Leftrightarrow AC \perp BD$$
