67th NMO Selection Tests for JBMO

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 $B^{\prime }$ be the foot of the altitude from $B$ and $C^{\prime }$ be the foot of the altitude from $C$.

Since $BC{C}^{\prime }{B}^{\prime }$ is a cyclic quadrilateral we have $\angle A{B}^{\prime }{C}^{\prime }\equiv \angle ABC$. Hence $△A{B}^{\prime }{C}^{\prime }\sim △ABC$, having the similarity ratio $\frac{A{C}^{\prime }}{AC}=\cos (\widehat{BAC})=\frac{1}{2}$. It follows that the similarity ratio is the same with the ratio of the diameters of the circumcircles of triangles $A{B}^{\prime }{C}^{\prime }$ 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.