2024 CMO

Proof. Let $K$ be the midpoint of $ID$, then $LN$ passes through $K$. Next, we prove that both $EF$ and $PQ$ pass through $K$. Let $X$ and $Y$ be the reflections of $D$ across $E$ and $F$, respectively. Let $Z$ be the projection of $I$ onto $AC$. Then, we have $$CX=2DE+CD=AD+BD-AB+CD=AC+BC-AB=2CZ.$$ Thus, $CZ=ZX$. Combining this with $IZ\perp CX$, we know $IC=IX$. Therefore, $$2\angle IXC=2\angle ICX=\angle ACB=\angle BDC=2\angle FEC=2\angle YXC.$$ Thus, $I$, $Y$, and $X$ are collinear, which implies that $EF$ passes through $K$. Since $MN//AI$, we have $\angle PQD=\angle PMD=\angle IAD$. From $\angle DMJ=90^{\circ }$, we know that $DJ$ is the diameter of $\omega$, so $DQ\perp QJ$. Combined with $AL\perp LJ$, we conclude that $AI//DQ$. Let $T$ be the reflection of $D$ across $Q$. Then $ADTI$ forms an isosceles trapezoid. Since $KQ//IT$, we have $$\angle DQK=\angle DTI=180^{\circ }-\angle IAD=180^{\circ }-\angle PQD,$$ which implies that $PQ$ passes through $K$. Thus, the proof is complete.