Reflection of orthocenter about side midpoints is the diameter of circumcircle

Question

Let H be the orthocenter of ABC. Let Y the reflection over the midpoint of BC. Show that AY is a diameter of (ABC), the circunference that passes through ABC.

My take: Using the fact that HCYB is a parallelogram (since its diagonals cross in their respective midpoints), we can state that $\angle$BYC = $\angle$BHC which is 180º - $\angle$ BAC. Since $\angle$ BAC + $\angle$ BYC = 180º, ABYC is a cyclic quadrilateral and Y lies on the circle. But I'm unsure on how to prove AY is a diameter, or that (which is equivalente) $\angle$ ACY = $\angle$ ABY = 90º. I know this question: Reflection of orthocenter about side midpoints lies on circumcircle has an answer which solves this with the use of complex numbers, but I would prefer a geometric aproach if possible.

Problem taken from Euclidean Geometry in Matematical Olympiads, by Evo Chen.

Answer 1

This solution doesn't use anything fancy.
These relationships can be observed if you drew the accurate diagram several times, which is a good first step for olympiad problems.
Fill in the details.

1. Show that $Y$ lies on the circumcircle (done in statement).
2. Let $X$ be the reflection of $H$ across $BC$. Show that $X$ also lies on the circumcircle by (similar) angle chasing.
3. Show that $XY \parallel BC$.
4. Show that $AHX$ is a straight line.
5. Since $AH \perp BC$, hence $\angle AXY = 90^ \circ$.
6. Hence $AY$ is the diameter of the circumcircle.

Answer 2

This solution follows from the fact that $H$ is the center of homothety (expansion) from the nine-point circle to the circumcenter. (You don't need to know this fact to understand the solution. In fact, this solution establishes this fact.)
It's a natural way to set up the question, and possibly how it was set.

Let $D$ be the midpoint of $BC$.
Let $P$ be the foot of the perpendicular from $A$ to $BC$.
Let $T$ be the midpoint of $AH$.
Recall that $DPT$ are 3 points of the nine-point circle, whose center $N$ is the midpoint of $OH$. $DNT$ is a diameter of the circle (since $\angle DPT = 90^\circ$).

By considering triangle $AOH$, it follows that $TN \parallel AO$ by midpoints.
By considering triangle $HOY$, it follows that $ND \parallel OY$ by midpoints.
Since $TND$ is a straight line, hence $AOY$ is a straight line, so $AY$ is the diameter of the circumcircle (given that $Y$ lies on the circumcircle).