Problem 1749

Solution:

Without loss of generality assume that $\angle A D C \geq 90^{\circ}$. Let $O$ denote the circumcenter of triangle $A B C$ and $K$ the orthocentre of triangle $O_{1} O_{2} D$. We shall first show that the points $O$ and $K$ lie on the circumcircle of triangle $A O_{1} O_{2}$. Note that circumcircles of triangles $A B D$ and $A C D$ pass through the points $A$ and $D$, so $A D$ is perpendicular to $O_{1} O_{2}$ and, triangle $A O_{1} O_{2}$ is congruent to triangle $D O_{1} O_{2}$. In particular, $\angle A O_{1} O_{2}=\angle O_{2} O_{1} D=\angle B$ since $O_{2} O_{1}$ is the perpendicular bisector of $A D$. On the other hand since $O O_{2}$ is the perpendicular bisector of $A C$ it follows that $\angle A O O_{2}=\angle B$. This shows that $O$ lies on the circumcircle of triangle $A O_{1} O_{2}$. Note also that, since $A D$ is perpendicular to $O_{1} O_{2}$, we have $\angle O_{2} K A=90^{\circ}-\angle O_{1} O_{2} K=\angle O_{2} O_{1} D=\angle B$. This proves that $K$ also lies on the circumcircle of triangle $A O_{1} O_{2}$.

Therefore $\angle A K O=180^{\circ}-\angle A O_{2} O=\angle A D C$ and hence $O K$ is parallel to $B C$.