<s>[INST] 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.
 [/INST] 
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 $=0 \ \implies E=F \ \implies AC\perp BD$
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