22nd Korean Mathematical Olympiad Final Round

Let $K^{\prime }$ be the intersection of the line $CD$ and the line passing through $E$ and parallel to the line $CL$. Since $\angle EBD=\angle DCL=\angle E{K}^{\prime }D$, we have that the four points $B$, $D$, $E$ and $K^{\prime }$ are on one circle, say $O_1$. Since $A{E}^2=A{C}^2=AD\cdot AB$, we have that the line $AE$ is tangent to the circle $O_1$ at $E$.

Now consider the homothety $H$ which sends $O_1$ to $O$. The center of the homothety $H$ is the point $F$ and $H$ maps points $E$ and $O_1$ to $C$ and $O$, respectively. And also $H$ maps the point $K^{\prime }$ to the point $L$. So three points $F$, $K^{\prime }$ and $L$ are collinear and thus we have $K^{\prime }=K$. This completes the proof. $\square$