Let $H$ and $O$ be the orthocenter and the circumcenter of $ABC$, $R$ the radius of the circumcenter; then $OH$ meets the line $AB$ at $D$ and the line $AC$ at $E$. Let $B'$ be the foot of the altitude from $B$ and $C'$ be the foot of the altitude from $C$.

Since $BCC'B'$ is a cyclic quadrilateral we have $\angle AB'C' \equiv \angle ABC$. Hence $\triangle AB'C' \sim \triangle ABC$, having the similarity ratio $\frac{AC'}{AC} = \cos(\widehat{BAC}) = \frac{1}{2}$. It follows that the similarity ratio is the same with the ratio of the diameters of the circumcircles of triangles $AB'C'$ and $ABC$, so $\frac{AH}{2R} = \frac{1}{2}$, which leads to $AH = R = AO$. (1)

It is known that the rays ($AH$ and ($AO$ are isogonal, so $\angle BAO \equiv \angle CAH$. (2) From (1) it results that $\angle AOH \equiv \angle AHO$, so $\angle AOD \equiv \angle AHE$. Using (1) and (2), it follows that triangles $AOD$ and $AHE$ are congruent, so $AD = AE$ and the conclusion follows.