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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.

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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?).

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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: $2a^2+2b^2=e^2+f^2$.

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  • $\begingroup$ Thank you for the response! $\endgroup$ – nicoledobreva May 4 '20 at 18:54
  • $\begingroup$ Can you help me to show that this equality holds always, regardless of whether the diagonals are perpendicular or not? $\endgroup$ – nicoledobreva May 4 '20 at 18:54
  • $\begingroup$ As I wrote in the post, $KLMN$ is a parallelogram by Varignon's Theorem. If we use the fact you wrote, $2MN^2+2ML^2=KM^2+LN^2$. $\endgroup$ – nicoledobreva May 4 '20 at 18:56
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    $\begingroup$ Yeah and now use $2MN=AC$ etc. $\endgroup$ – timon92 May 4 '20 at 18:57
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    $\begingroup$ $AB^2+CD^2=BC^2+DA^2$ iff the diagonals are perpendicular. $\endgroup$ – user May 4 '20 at 19:27

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