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David-Vieta wrote:

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Thanks.

My solution during the contest: Let H' be the orthocenter of AEF. Clearly AD bisects segment EF. Let midpoint of EF be G. EHFH' is a parallelogram, so HG=H'G. Angle chasing gets HD//AH', and the rest is easy.

Attachments:

In fact there are more in this diagram. In other words, I gave a much more complicated solution. For example, Let $O$ be the center of $\odot ABC$. Then $\triangle AHD\sim\triangle AOM$.

In acute triangle $ABC$. $AB>AC$. $H$ is the orthocenter. $M$ is midpoint of $BC$ and $AD$ is the symmedian line. Prove that if $\angle ADH= \angle MAH$, $EF$ bisects segment $AD$.