SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Suppose that $KA$ cuts $PF$ at $M$, $KB$ cuts $PG$ at $N$. By angle chasing, we have $$\angle IEF=\angle BAI=\angle FAI,$$ then $AIFE$ is cyclic. In addition, we get $$\begin{array}{rl}\angle AMF=\angle KAI& =\angle KAB+\angle BAI\\ & =\angle AEI+\angle FEI=\angle AEF\end{array}$$ then $AIFM$ is cyclic.



Summarily, points $A,I,F,E,M$ lie on a circle. And similarly, points $B,I,G,D,N$ also lie on a circle. We have known that $DE$ is perpendicular bisector of $CI$, so it is easy to see that $IFCG$ is a rhombus. From that above, we get $$\angle AMI=\angle AEI=\angle KAB,$$ then $IM\parallel AB$. Similarly $IN\parallel AB$, implies that $M,N,I$ are collinear. But from $PF\parallel AD$, $PG\parallel BE$, we have $AIFM$ and $BIGN$ are isosceles trapezoids, so $$\begin{array}{c}AM=IF=IG=BN,\end{array}\text{\hspace{1em}(1)}$$ then $ABNM$ is also an isosceles trapezoid, which means that it is cyclic.

In the other hand, we get $$\begin{array}{c}\angle PFG=\angle MFE=\angle MIE=\angle BIN=\angle ING,\end{array}\text{\hspace{1em}(2)}$$ so $FGNM$ is cyclic.

From (1) and (2), we have $PK$ is the radical axis of two circles $(AIE)$ and $(BID)$. Now on, we can easily take the conclusion of this problem. $\square$