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 $\Delta$$ABC$ is $O$, and the orthocenter of $\Delta$$O_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 :)]
Best Answer
I proceded starting from your suggestions and then by the following path. You can use your result to state that \begin{equation} \angle O_2OH \cong\angle O_2AH \cong \angle O_2DH,\tag{1}\label{eq:cong1} \end{equation} 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 \begin{equation} \angle O_2OM_1 + \angle ACB \cong \pi,\tag{2}\label{eq:sum1} \end{equation} 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 \begin{equation} \angle ACB + \angle O_2OH \cong \frac{\pi}{2}, \end{equation} which will lead to the thesis.