In the drawing, the circumference has center $O$ and radius $r$. If $MP=a$ and $PN=b$, and the sum of the arcs $AM$ and $NB$ is $90º$, then $a^2+b^2=?$
The only thing i discover is that the interior angles are $135º$ and $45º$, but nothing else. Any hints?
If you reflect $MN$ across $AB$ you get new chord $M'N'$ (which goes through $P$). Since $\angle MPN' = \angle MON'= 90^{\circ}$ we have cyclic quadrilateral $MPON'$ with diameter $MN'$. So by Pythagoras we have $$a^2+b^2 = MN'^2 =r^2+r^2 =2r^2$$