# Geometry involving circumcenters

In an acute-angled triangle $$ABC$$, a point $$D$$ lies on the segment $$BC$$. Let $$O_1$$,$$O_2$$ denote the circumcentres of triangles $$ABD$$ and $$ACD$$, respectively. Prove that the line joining the circumcentre of triangle $$ABC$$ and the orthocentre of triangle $$O_1O_2D$$ is parallel to $$BC$$.

Supposing that the circumcentre of $$\DeltaABC$$ is $$O$$, and the orthocenter of $$\DeltaO_1O_2D$$ is H, I could prove that $$A,O_1,O,H,O_2$$ lie on a circle. After that I cannot figure out how to do. Please help.

[Any other better solution is also welcome :)]

I proceded starting from your suggestions and then by the following path. You can use your result to state that $$$$\angle O_2OH \cong\angle O_2AH \cong \angle O_2DH,\tag{1}\label{eq:cong1}$$$$ where the first congruence is due to the fact that the angles are subtended by the same chord $$O_2H$$, and the second congruence is due to the fact that triangle $$AO_2D$$ is isosceles. Call then $$M_1$$ the middle point of $$BC$$ and $$M_2$$ the middle point of $$AC$$. Note that $$$$\angle O_2OM_1 + \angle ACB \cong \pi,\tag{2}\label{eq:sum1}$$$$ which can be obtained by adding up the internal angles of quadrilateral $$CM_2OM_1$$. Now subtract and add $$\angle O_2OH$$ on the LHS of \eqref{eq:sum1}, getting $$\begin{eqnarray} \angle O_2OM_1 -\angle O_2OH+ \angle ACB + \angle O_2OH&\cong& \pi\\ \angle HOM_1 + \angle ACB + \angle O_2OH \cong \pi. \end{eqnarray}$$ Finally, by using \eqref{eq:cong1} and properties of the circumcenter $$O_2$$, show that $$$$\angle ACB + \angle O_2OH \cong \frac{\pi}{2},$$$$ which will lead to the thesis.