How to prove that perpendicular bisector of line joining altitudes also bisects the side?

Question

Let ${D,E,F}$ be the feet of the altitude from ${A,B,C}$ in a ${\triangle{ABC}}$. Prove that the perpendicular bisector of ${EF}$ also bisects ${BC}$.

Answer 1

The only important thing is that $\square BCEF$ lies on a circle, since both $E$ and $F$ look at $\overline{BC}$ at right angle. But $\overline{BC}$ is a diameter while $\overline{EF}$ is just a line segment connecting two points on a circle. Now the perpendicular bisector to that would intersect any diameter in the center of the circle, i.e. the midpoint of the diameter.