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

Question

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

I've tried proving this using congruence but to no avail. How should I proceed? Please help.

Thanks.

Answer 1

It's fairly easy to notice that $EFCB$ is cyclic and moreover as $\angle CEB = \angle CFB = \frac{\pi}{2}$ we have that if $M$ is the midpoint of $BC$ it's also the center of the circumcircle of $EFCB$. In this circle $EF$ is a chord and it's well-known that the bisector of any chord passes through the center of the circle. Because of all this the bisector of $EF$ passes through $M$, the midpoint of $BC$.