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](https://math.stackexchange.com/questions/132321/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.