Belarusian Mathematical Olympiad

Let $\angle APD=\angle BPC=\phi$. Then $\angle PKA=\angle PLB=\frac{\pi -\phi }{2}$ since the triangles $APK$ and $BPL$ are isosceles. Therefore, $MK=ML$.



From the equalities $AP=PK$, $BP=PL$ and the power of a point theorem it follows that $PK\cdot PC=AP\cdot PC=BP\cdot PD=PL\cdot PD$. Hence, $\frac{PK}{PL}=\frac{PD}{PC}$. Further, $\angle KPD=\angle LPC=\frac{\phi }{2}$. Therefore, the triangles $KPD$, $LPC$ are similar, so $\angle PKD=\angle PLC$. Hence, $NK=NL$. Now we have $MK=ML$ and $NK=NL$. It follows that $M$ and $N$ lie on the perpendicular bisector of the segment $KL$, which gives the required statement.