23rd Junior Turkish Mathematical Olympiad

Let $AM$ and $DE$ intersect at $N$, and $F$ be the foot of the perpendicular line from $O$ to $AC$. Since $AO\perp DE$ and $\angle BAO=90^{\circ }-\angle ACB$ we get $\angle ALD=\angle AED=\angle ACB$ and hence the points $K$, $C$, $D$, $L$ are concyclic. Using power equations we obtain $$AL\cdot AK=AD\cdot AC.(1)$$

Since $F$ is the midpoint of $[AC]$ and $N$ is the midpoint of $[AM]$, we get $$AC=2AF\text{and}AM=2AN.(2)$$

On the other hand, $\angle DNO=\angle OFD=90^{\circ }$ implies that the points $D$, $N$, $O$, $F$ are also concyclic. Using power equations again, we get $$AN\cdot AO=AD\cdot AF.(3)$$ Using (1), (2) and (3), we conclude that $$AL\cdot AK=AD\cdot AC=2AF\cdot AD=2AN\cdot AO=AM\cdot AO$$ which shows that the points $K$, $L$, $M$, $O$ are concyclic.