SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Let $AD$ be the altitude of triangle $ABC$, $A{A}^{\prime }$ be the diameter of $(O)$. $AO$ meets $EF$ at $N$. Let $L$ be the reflection of $A$ through $N$.



We have $AO\perp EF$ then quadrilateral $NEC{A}^{\prime }$ is cyclic. Then $$AO\cdot AL=\frac{1}{2}A{A}^{\prime }\cdot 2AN=A{A}^{\prime }\cdot AN=AE\cdot AC=AH\cdot AD=AK\cdot AM.$$ From this we obtain $KOLM$ is cyclic.

On the other side, $AQ\parallel EF$ and $P$ is the midpoint of $QT$ then $ALTQ$ is a trapezoid with its midline $PN$. We get $$\angle LTC=\angle EPC=\angle HAO=\angle MOL.$$ Hence $OMTL$ is cyclic, which follows that $O,K,M,T,L$ are concyclic or $K$ lies on circle with diameter $OT$.

In other words, $\angle OKT=90^{\circ }$. $\square$