A question on geometry? I wanted to know, given quadrilateral ABCD such that $AB^2+CD^2=BC^2+AD^2$ , prove that $AC⊥BD$ .
Help. 
Thanks.
 A: Here's a solution using vector geometry.
According to the Problem,
$$(\vec{B}-\vec{A})^{2}+(\vec{D}-\vec{C})^{2}=(\vec{C}-\vec{B})^{2}+(\vec{D}-\vec{A})^{2}$$
$$\rightarrow A^{2}+B^{2}+C^{2}+D^{2}-2\vec{A}\cdot \vec{B}-2\vec{C}\cdot\vec{D}=A^{2}+B^{2}+C^{2}+D^{2}-2\vec{C}\cdot \vec{B}-2\vec{D}\cdot\vec{A}$$
 $$\rightarrow(\vec{A}-\vec{C})\cdot (\vec{B}-\vec{D})=0$$
which implies that $AC\perp BD$
A: Hints:
I'll do it with analytic geometry: place the vertices at
$$D=(0,0)\;,\;C=(x,0)\;,\;A=(a,b)\;,\;B=(\alpha,\beta)$$
Then we get that
$$AB^2+CD^2=(a-\alpha)^2+(b-\beta)^2+x^2=b^2+(\alpha-x)^2+a^2+b^2=BC^2+AD^2\iff$$
$$a\alpha+b\beta=x\alpha\iff \frac\alpha\beta=\frac b{x-a}$$
Now check what the slopes of $\,AC\,,\,BD\,$ are and what's the condition they must fulfill in order to get $\,AC\perp BD\;$ ...
A: 
Here is Geometry method:
$AB^2=BE^2+AE^2, DC^2=DF^2+CF^2, BC^2=BE^2+CE^2,AD^2=AF^2+DF^2$
$AB^2+CD^2=BC^2+AD^2 \implies BE^2+AE^2+DF^2+CF^2=BE^2+CE^2+AF^2+DF^2$
$\implies AE^2+CF^2=CE^2+AF^2 \implies AE^2- AF^2=CE^2-CF^2$ .....<1>
$AE \ge AF, CE \le CF $ so for <1>:
LHS $\ge 0$, RHS $ \le 0$, the only condition is both sides $=0 \ \implies E=F \ \implies AC\perp BD$
A: Let $AC,BD$ intersect at $O$.  Let $ (AO,BO,CO,DO)= (p,q,r,s) $ respectively.
$ p^2 + q^2= AB^2 , r^2 + s^2 = CD^2$
$ q^2 + r^2 = BC^2 , p^2 + s^2= AD^2 $
and add up each line.
