Prove that the symmetric point of orthocenter $H$ of a triangle with respect to the midpoint of any side resides on circumcircle. 
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...
 A: 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 $ABC$.
A: Use that BHCK is a quadrilateral whose diagonals intersect in the middle. So is a parallelogram. Then show that BHCK is inscribed.
A: A complex number proof. Without loss of generality we may assume that $A,B,C$ are complex numbers with the unit circle of $\mathbb{C}$ centered at the origin $O$. Then  $K=-C$, $H=A+B+C$ and
$$\frac{H+K}{2}=\frac{A+B}{2}=M.$$
As a reference I suggest Bashing Geometry with Complex Numbers by E. Chen.
