Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.1 National olympiad Prove it

2. In the acute triangle $A B C, M$ is a point in the interior of the segment $A C$ and $N$ is a point on the extension of the segment $A C$ such that $M N=A C$ . Let $D$ and $E$ be the feet of the perpendiculars from $M$ and $N$ onto the lines $B C$ and $A B$ respectively. Prove that the orthocentre of $\triangle A B C$ lies on the cicumcircle of $\triangle B E D$ .

Solution

2. Let $K$ be the point of intersection of $M D$ and $N E$ . It is easy to see that the circle with diameter $B K$ is the circumcircle of $\triangle B E D$ . As $A H$ is parallel to $M K$ and $C H$ is parallel to $N K$ , we have $\angle H A C=\angle K M N$ and $\angle A C H=\angle M N K$ . Since $A C=M N$ , we thus have $\triangle A H C$ is congruent to $\triangle M K N$ .

Therefore the distance from $K$ onto $A C$ equals to the distance from $H$ onto $A C$ . But $H$ and $K$ are on the same side with respect to the line $A C$ , it follows that $H K$ is parallel to $A C$ . Therefore $H K$ is perpendicular to $B H$ and $H$ lies on the circle with diameter $B K$ circumscribing about $\triangle B E D$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.