National Math Olympiad

Denote the angle $\angle ABC$ by $\beta$. The line $KL$ is parallel to the altitude to the side $BC$, so $KL$ is perpendicular to $BC$. Hence, $KLB$ is a right triangle and $\angle KLB=\frac{\pi }{2}-\beta$. We get $\angle KLN=\frac{\pi }{2}-\beta$. Since $KLMN$ is a cyclic quadrilateral, we have $\angle KMN=\angle KLN=\frac{\pi }{2}-\beta$.



In the quadrilateral $KDMC$ we $A$ have $\angle DKC+\angle DMC=\frac{\pi }{2}+\frac{\pi }{2}=\pi$, so $KDMC$ is also a cyclic quadrilateral. We conclude that $\angle DCK=\angle DMK=\angle NMK=\frac{\pi }{2}-\beta$.

Let $E$ be the point where the line $CD$ intersects the segment $AB$. As we have shown we have $\angle ECB=\frac{\pi }{2}-\beta$ and $\angle EBC=\beta$, so $\angle BEC=\frac{\pi }{2}$. The line through the points $C$ and $D$ is perpendicular to the segment $AB$, so it must also contain the point $H$. Thus, $C$, $D$ and $H$ are collinear.