Let $H$ be the orthocentre of an acute $\Delta ABC$, as in figure. Let $X$ be the reflection of $H$ over $\overline{BC}$ and $Y$ be the reflection of $H$ over the midpoint of $\overline{BC}$. Show that $X$ and $Y$ lie on the circumcircle of $\Delta ABC$.
To prove that $X$ and $Y$ lie on the circumcircle of $\Delta ABC$, I tried showing that $AXYC$ and $ABXY$ are both cyclic which would make $A, B, C, X, Y$ concyclic and since the circumcircle is unique for a given triangle, $X$ and $Y$ must lie on it (follows from the points being concyclic).
I doubt this is sufficient as a proof for this problem. Can someone guide me? I would like to continue with this proof unless it's not convenient to prove by these arguments.
I will take it that it is already established that points $A$, $C$, $X$ and $B$ are concyclic. Now, we shall construct the perpendicular bisector of $BC$, which we will call $D$. Reflecting points $H$ and $X$ around line $D$ give $H'$ and $Y$ respectively. (I will leave you to prove that reflecting $X$ around line $D$ gives $Y$. You will need to use $H'$. )
As $BC$ is a chord of the circumcircle of $\triangle ABC$, its perpendicular bisector $F$ is a diameter of the circle. Thus, $A$, $B$, $C$, $X$ and $Y$ are concyclic. (The reflection of a point lying on the circumference of a circle upon its diameter lies on that circle itself.)