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Slovenia — National Math Olympiad

Geometry Difficulty 6.7 National Olympiad Prove it Slovenia

Let $H$ be the orthocentre of the acute triangle $ABC$ and let $D$ be a point inside the triangle $ABH$ . A line that passes through the point $D$ and is parallel to the line $AH$ intersects the segments $BC$ and $AB$ at $K$ and $L$ . A line that passes through the point $D$ and is parallel to the line $BH$ intersects the segments $AC$ and $AB$ at $M$ and $N$ . Prove that $C, D$ and $H$ lie on the same line if $K, L, M$ and $N$ lie on the same circle.

Solution

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.

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