# Show that the diagonals in a quadrilateral are perpendicular

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?).

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: $$2a^2+2b^2=e^2+f^2$$.
• As I wrote in the post, $KLMN$ is a parallelogram by Varignon's Theorem. If we use the fact you wrote, $2MN^2+2ML^2=KM^2+LN^2$. – nicoledobreva May 4 '20 at 18:56
• Yeah and now use $2MN=AC$ etc. – timon92 May 4 '20 at 18:57
• $AB^2+CD^2=BC^2+DA^2$ iff the diagonals are perpendicular. – user May 4 '20 at 19:27