# 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 :)]

• Please add a diagram. – Anubhab Ghosal Jan 1 at 16:29
• Well, the original question did not provide a diagram, the reader had to draw the diagram himself/ herself because there may be variations. – Yellow Jan 1 at 16:50
• 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. – Anubhab Ghosal Jan 1 at 16:55
• 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! – Yellow Jan 1 at 17:02
• 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). – Anubhab Ghosal Jan 1 at 19:23

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.