Let $O$ be a center of circle which passes through vertices of quadrilateral $ABCD$, which has perpendicular diagonals. Prove that sum of distances of point $O$ to sides of quadrilateral $ABCD$ is equal to half of perimeter of $ABCD$.
Let $H$ be the intersection of the diagonals and $M,N$ be the midpoints of $AB$ and $CD$ respectively. Just use some angle chasing to prove $OMHN$ is a parallelogram.
With some easy angle chasing we get that (without the loss of generality) that ∠ABC=∠ADC=90,
now we have that O is the midpoint of AC.
Now we see that O is perpendicular to the midpoints of AB BC CD and AD
Let M N P Q be the midpoints of AB BC CD AD
we have that OM=BC/2 ON=BC/2 OP=AD/2 OQ= CD/2
The proof is finished