Let $M,N,K$ and $L$ be the midpoints of the sides $AB,BC,CD$ and $AD$ of the quadrilateral $ABCD$. Show that $AC$ and $BD$ are perpendicular iff $AC^2+BD^2=2MK^2+2NL^2$. I am supposed to solve the problem using the Pythagorean theorem.
First, let us show that if $AC \perp BD$, then $AC^2+BD^2=2MK^2+2NL^2$ (this is called necessity, right?). $KLMN$ is a parallelogram by Varignon's Theorem. I was able to show that in a quadrilateral $ABCD$ the diagonals are perpendicular iff $AB^2+CD^2=AD^2+BC^2$. I am not sure if we can use this here.
Second, we should show that if $AC^2+BD^2=2MK^2+2NL^2$, then $AC\perp BD$ (this is called sufficiency, right?).
Best Answer
I think the problem is false. This equality holds always, regardless of whether $AC\perp BD$ or not. To show this you can use the following fact: in a parallelogram with sides $a,b$ and diagonals $e,f$ the following holds: $2a^2+2b^2=e^2+f^2$.