MMO2025 Round 4

Since $\angle BEC=\angle BFC=90^{\circ }$, points $E$ and $F$ lie on circle $\omega$.

We observe that: $$\angle LAF=\angle DAB=\angle FCB=\angle FPL,$$ so quadrilateral $AFLP$ is cyclic; denote its circumcircle by ${\omega }_1$. Similarly, we get that $AQLE$ is cyclic; denote its circumcircle by ${\omega }_2$. Now consider the radical axes of these three circles:

- The radical axis of $\omega$ and ${\omega }_1$ is line $PF$. - The radical axis of $\omega$ and ${\omega }_2$ is line $EQ$. - The radical axis of ${\omega }_1$ and ${\omega }_2$ is line $AL$.

By the Radical Axis Theorem, these three radical axes are concurrent. Therefore, the lines $AD$, $QE$, and $PF$ meet at a single point.