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China Hong Kong Mathematical Olympiad

Geometry Difficulty 4.5 AIME Prove it Hong Kong

Let $ABCD$ be a cyclic quadrilateral. $K, L, M, N$ are midpoints of sides $AB, BC, CD$ and $DA$ respectively. Prove that the orthocentres of triangles $AKN, BKL, CLM, DMN$ are vertices of a parallelogram.

Solution

Let $H_1, H_2, H_3, H_4$ be the orthocentres of $\triangle AKN, \triangle BKL, \triangle CLM, \triangle DMN$
respectively, and let $O$ be the centre of $(ABCD)$ .
Since $K$ is the midpoint of $AB$ , we have $OK \perp AB$ . Thus, $OK // H_1N$ . Similarly, $ON // H_1K$ . This shows $H_1KON$ is a parallelogram. By symmetry, $H_4NOM$ is a parallelogram. This shows $H_1K = NO = H_4M$ and $H_1K // NO // H_4M$ . Therefore, $H_1KMH_4$ is a parallelogram. By symmetry, $KH_2H_3M$ is a parallelogram. This shows $H_1H_4 = KM = H_2H_3$ and $H_1H_4 // KM // H_2H_3$ . Therefore, $H_1H_2H_3H_4$ is a parallelogram.

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