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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.

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    $\begingroup$ 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. $\endgroup$
    – Calvin Lin
    Feb 19, 2023 at 13:34

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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 $\triangle ABC$.

We are done.

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