# Prove that $\angle BQC = 2 \angle BAC$

Let the acute triangle $$ABC$$ and the points $$M$$, $$N$$, $$P$$ the means of the sides $$BC$$, $$CA$$ and $$AB$$, respectively. We denote by $$Q$$ the orthocenter of the triangle $$MNP$$. Prove that $$\angle BQC = 2 \angle BAC$$.

I thought about the middle line theorem. We apply it $$3$$ times for each line of the triangle. We will get $$4$$ congruent triangles and two similar ones. From here I don't know what to do exactly. I have some ideas:

1. We show that the triangle $$ABC$$ or the median triangle are equilateral. If the triangle is equilateral, we can inscribe it in a circle, showing that $$Q$$ is the center of the circumscribed circle.
2. We show that $$A$$, $$Q$$, $$M$$ are collinear. If they are collinear $$PN\parallel BC$$ but $$MQ$$ perpendicular to $$PB$$ $$\implies$$ $$AM$$ perpendicular to $$BC$$. Analogous for the other sides $$\implies$$ $$Q$$ is the center of the circumscribed circle. We apply the angle property $$\implies$$ conclusion.
3. We show that $$Q$$ is the midpoint of $$PN$$ $$\implies$$ $$ANMP$$ rhombus $$\implies$$ continuation $$2$$

However, I don't know how to show the first part of each idea. Hope one of you can help me! Any ideas are welcome.

• 1/ Presumably you want the ANGLE BQC = 2 ANGLE BAC. 2/ If so, since this holds similarly for the other sides, this means that Q is the circumcenter of ABC (which is what you alluded to). Try proving this using angle chasing or similarity. Feb 19, 2023 at 13:34

Note that the altitude in $$\triangle MNP$$ perpendicular to $$PN$$, is perpendicular to $$BC$$ because $$PN$$ and $$BC$$ are parallel. Moreover $$M$$ is the midpoint; hence this altitude is indeed the bisector of $$BC$$. The same goes for the two other altitudes. Therefore,
$$Q$$ as the intersection of the three altitudes of $$\triangle MNP$$ is actually the intersection of the three bisectors of $$AB$$, $$AC$$, and $$BC$$.
So, $$Q$$ is both the the orthocenter of $$\triangle MNP$$ and the circumcenter of $$\triangle ABC$$.