Prove that the symmetric point of orthocenter $H$ of a triangle with respect to the midpoint of any side resides on circumcircle.

Question

Prove that the symmetric point of orthocenter $H$ of a triangle with respect to the midpoint of any side resides on the triangle's circumcircle.

In the above figure,it's sufficient to prove that triangle $BHM$ is equal to triangle $CKM$, but how? I tried to make use of the symmetric point of $H$ with respect to $BC$ which resides on circumcircle but failed...

Answer 1

The symmetric of $H$ with respect to the $BC$-side, say $J$, lies on the circumcircle of $ABC$ since $\widehat{BHC}+\widehat{BAC}=\pi$. If we consider the symmetric of $J$ with respect to the perpendicular bisector of $BC$ we get $K$, hence $K$ lies on the circumcircle of $ABC$, too, since the circumcircle of $ABC$ is symmetric with respect to the perpendicular bisector of $BC$.

As an alternative, from the statement it clearly follows that $BHCK$ is a parallelogram, since its diagonals $BC$ and $HK$ intersect at their midpoints. That gives $\widehat{CKB}=\widehat{BHC}=\pi-\widehat{BAC}$, hence $K$ lies on the circumcircle of $ABC$.

Answer 2

Use that BHCK is a quadrilateral whose diagonals intersect in the middle. So is a parallelogram. Then show that BHCK is inscribed.