$ABCD$ is a convex quadrilateral in which $AB^2+CD^2=BC^2+AD^2$. Prove by vectors that diagonals $AC$ and $BD$ are orthogonal to each other. 
$ABCD$ is a convex quadrilateral in which $AB^2+CD^2=BC^2+AD^2$. Prove by vectors that diagonals $AC$ and $BD$ are orthogonal to each other.

How should i prove it? I think Ptolemy's theorem is not working here.
Please help me.
 A: Let me try. Let $O$ be an intersection of $AC$ and $BD$
Vector method:
One has 
$$\begin{align}
\vec{AO} + \vec{OB} = \vec{AB} \implies AO^2 + OB^2 + 2\vec{AO}\cdot\vec{OB} &= AB^2.\\
BO^2 + OC^2 + 2\vec{BO}\cdot\vec{OC} &= BC^2.\\
CO^2 + OD^2 + 2\vec{CO}\cdot\vec{OD} &= CD^2.\\
DO^2 + OA^2 + 2\vec{DO}\cdot\vec{OA} &= DA^2.
\end{align}$$
From $AB^2+CD^2= BC^2+DA^2$, we have
$$\begin{align}
\vec{AO}\cdot\vec{OB} + \vec{CO}\cdot\vec{OD} &= \vec{BO}\cdot\vec{OC} + \vec{DO}\cdot\vec{OA} \\
\vec{AO}\cdot\vec{OB} + \vec{CO}\cdot\vec{OD} + \vec{OB}\cdot\vec{OC} + \vec{OD}\cdot\vec{OA} &= \vec{0} \\
\vec{AC}\cdot\vec{OB} + \vec{CA}\cdot\vec{OD} &= \vec{0} \\
\vec{AC}\cdot\vec{DB} &= \vec{0}
\end{align}$$
Then $AC \perp BD$.
A: Since
$$\vec{AC}=\vec{AB}+\vec{BC}=\vec{AD}+\vec{DC}$$
we have
$$\vec{AB}-\vec{DC}=\vec{AD}-\vec{BC}.$$
Square both sides and use $AB^2+CD^2=AD^2+BC^2$, we get
$$\vec{AB}\cdot\vec{DC}=\vec{AD}\cdot\vec{BC}.$$
That means
$$(\vec{AB}+\vec{BC})\cdot\vec{DC}=(\vec{AD}+\vec{DC})\cdot\vec{BC},$$
or
$$\vec{AC}\cdot \vec{DC}=\vec{AC}\cdot\vec{BC}.$$
So $\vec{AC}\cdot\vec{BD}=0$.
