, 2009

Extend $AK$ to meet the circum-circle of $ABC$ in $D$, and join $MD$. Let $P$ be the point of intersection of $AK$ and $CL$. Observe that $\angle DMK = 90^\circ$ and $D, M, O$ are collinear. We show that $DMK$ is similar to $CPK$, which proves that $CL$ is perpendicular to $AK$. It is sufficient to prove that $KD/KC = KM/KP$. But $AK \cdot KD = BK \cdot KC$, which gives $KD/KC = BK/AK$. Thus we need to prove that
$$\frac{KM}{KP} = \frac{BK}{AK}.$$

Since $CL$ is a transversal in the triangle $AMK$. Menelaus' theorem gives
$$\frac{PA}{KP} = \frac{AL \cdot MC}{LM \cdot CK}.$$
But $KL$ is parallel to $CA$, so that $AL/LM = CK/KM$. This implies that
$$\frac{PA}{KP} = \frac{CM}{KM}.$$
Thus
$$\frac{AK}{KP} = \frac{CM + MK}{KM}.$$
All we need to show is $BK = CM + MK$. Since $BM = MC$, the result follows.