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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 :)]

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  • $\begingroup$ Please add a diagram. $\endgroup$ – Anubhab Ghosal Jan 1 at 16:29
  • $\begingroup$ Well, the original question did not provide a diagram, the reader had to draw the diagram himself/ herself because there may be variations. $\endgroup$ – Yellow Jan 1 at 16:50
  • $\begingroup$ This is not a site like Brilliant or AOPS. Here, the questioner must show effort and then the answer is to help him/her to complete the solution. One has to draw a diagram to solve the problem. Hence, the question is easier to comprehend with a diagram. As for the fact that multiple configurations are possible, draw any possible configuration. $\endgroup$ – Anubhab Ghosal Jan 1 at 16:55
  • $\begingroup$ I do understand, @Anubhab Ghosal. I am not simply posting questions or asking doubts. I do write whatever I have done as a part of my attempt to solve the problem. But please see, I registered with this website just yesterday and I am yet to explore and learn more on this website. So maybe in my future posts I will try my best to post a figure, along with the question, too. Anyways, thanks for the suggestions! $\endgroup$ – Yellow Jan 1 at 17:02
  • $\begingroup$ You are welcome. Welcome to Math.SE. Enjoy! (I am not sure if I am the right person to welcome you as I myself joined only a month back. :P). $\endgroup$ – Anubhab Ghosal Jan 1 at 19:23
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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.

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