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My thoughts here are to show that the 3 reflected points are all equidistant from the circumcenter and hence they uniquely determine a circle (the circumscribed circle). However, I'm not sure how to find a relation between the orthocenter and the midpoints of the sides.

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  • $\begingroup$ Could you tell me where you got this question? $\endgroup$
    – Cataline
    Sep 10, 2015 at 20:56
  • $\begingroup$ The same question (Lemma 1.17) is also in EGMO, Evan Chen $\endgroup$
    – sato
    Jul 28, 2021 at 5:10

1 Answer 1

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Suppose $D$ is the reflected of orthocenter $H$ around the midpoint of $BC$. Then it is not difficult to prove that $BDCH$ is a parallelogram and $\angle BDC = \angle BHC = \pi - \angle BAC$, which is a sufficient condition for $ABCD$ to lie on a circle.

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