In acute $\Delta ABC$, $O$ is circumcentre. $M,N$ on $AB,AC$, s.t. $O$ lies on $MN$. $D,E,F$ are midpoints of $MN,BN,CM$. Prove O,D,E,F are concyclic.

Question

In acute $\Delta ABC$, $O$ is the circumcentre. $M,N$ lie on $AB,AC$ respectively, s.t. $O$ lies on $MN$. $D,E,F$ are midpoints of $MN,BN,CM$ respectively. Prove that $O,D,E,F$ are concyclic.

I was able to prove that $DE||AB$ and $DF||AC$, and that $\angle EDF = \angle BAC$. Thus it remains to prove that $\angle EOF = \angle BAC$. However, I can't seem to get a relation between $\angle EOF$ and other angles.

Moreover, on Geogebra, I found that $\Delta AMN \sim \Delta OFE$, but I can't prove it.

[This was given to me as an exercise by my teacher.]

Answer 1

Let $B',C'$ be points on the circumcircle so that $BB',CC'$ are diameters. Then $OE\parallel B'N$, $OF\parallel C'M$. Now $B'N, C'M$ meets at some point $X$ that is on the circumcircle:

Proof: The intersections $C'X\cap AB=M, BB'\cap CC'=O, AC'\cap B'X=N$ are collinear so the six points $A,B,C,B',C',X$ are conconic by Braikenridge–Maclaurin theorem. But $A,B,C,B',C'$ are all on the circumcircle of $\triangle ABC$, so $X$ is too. QED

Now we angle chase: \begin{align*} \angle BAC&=\angle BAX+\angle XAC\\ &=\angle BB'X+\angle XC'C\\ &=\angle BOE+\angle COF \end{align*} and $\angle BOC=2\angle BAC$, so \begin{align*} \angle EOF&=\angle BOC-\angle BOE-\angle COF\\ &=\angle BAC \end{align*} and we are done.