lssl wrote: Let $O$ be the circumcenter of triangle $ABC$ , a line passes through $O$ intersects sides $AB,{\color{red}AC}$ at points $M,N$ , $E$ is the midpoint of $MC$ , $F$ is the midpoint of $NB$ , prove that : $\angle FOE= \angle BAC$ Typo fixed. Let $D$ be midpoint of $MN$ $\Longrightarrow$ $DE \parallel AC, DF \parallel AB$ $\Longrightarrow$ $\angle EDF = \angle CAB$. $OFED$ is cyclic, see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=291269 $\Longrightarrow$ $\angle EOF = \angle EDF = \angle CAB$.

Assume $BO$ meets $\odot O$ at another point $J$, $CO$ meets $\odot O$ at another point $K$, $JN$ meets $\odot O$ at another point $P$, $PK$ intersects $AB$ at $M'$. For the six concyclic points of $K,A,J,C,P,B$, by the pascal's theorem we have $N(PJ\cap AC),O(BJ\cap CK),M'(AB\cap PK)$ are collinear. On the other side, line $ON$ intersects side $AB$ at point $M$, hence $M=M'$ which means $K,M,P$ are colliner. Easily we know $OF$ is the median of $\triangle BJN$, therefore $OF \parallel JP$. Similarily we have $OE \parallel KP$. Hence $\angle EOF=\angle KPJ$. Since $BJ$ is the diameter of $\odot O$, thus $JC\perp BC$. Similarily $KB\perp BC$. Hence $JC\parallel KB$, then arc $\widehat{KAJ}$=arc $\widehat{BPC}$, so $\angle A=\angle KPJ$. Eventually we got $\angle EOF=\angle A$.

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Moving Points Define point $F^*$ on the $B$-midline of $\triangle ABC$ such that $\angle EOF^* = \angle A$. Animate $M$ along $AB$. Obviously $M\to N$ is projective. Moreover, $M\to E$ and $N\to F$ are projective as they are homotheties. $E\to F'$ are also projective as it's rotation around $O$. Hence $F\to F^*$ is projective and it suffices to show $\angle FOE = \angle BAC$ for three different positions. When $O$ is midpoint of $MN$, we get $OF\parallel AB$ and $OE\parallel AC$, done. When $M=B$, we get $OE\perp BC$ thus $\angle EOF = \angle BOE = \angle A$, done. The case $N=C$ can be handled similarly so we are done.

Butterfly Let \(M', N'\) respectively be the reflection of \(M,N\) in point \(O.\) By the converse of butterfly theorem \(CM' \cap BN' \in (O).\) But clearly, \(CM' || OE\) and \(BN' || OF \Rightarrow \angle EOF = \angle A\).

Impressively simple. I just link to the solution above this pdf for someone (like me) who don't know about butterfly theorem in its general form.

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perhaps we can use Pascal to simplify the process

This is Iran 2004. We will use the following point labeling: Quote: Let $ABC$ be a triangle, and $O$ the center of its circumcircle. Let a line through the point $O$ intersect the lines $AB$ and $AC$ at the points $X$ and $Y$, respectively. Denote by $M$ and $N$ the midpoints of the segments $BY$ and $CX$, respectively. Prove that $\measuredangle MON=\measuredangle BAC$. We will use complex numbers with $(ABC)$ as the unit circle and $\ell$ as the real line (rotate appropriately). Set $O=0, A=a,B=b,C=c$ and let $\ell$ intersect $(ABC)$ at points $P=-1,Q=1$. By Lemma 8 in https://web.evanchen.cc/handouts/cmplx/en-cmplx.pdf, $X=\frac{a+b}{ab+1},Y=\frac{a+c}{ac+1}$ so we clearly have $M=m=\frac{b+y}2=\frac{a+b+c+abc}{2(ac+1)},N=n=\frac{c+x}2=\frac{a+b+c+abc}{2(ab+1)}$. To show that $\measuredangle BAC=\measuredangle MON$, we need to show $\text{arg~}\frac{a-b}{a-c}=\text{arg~}\frac{o-m}{o-n}$ or equivilently $\frac{o-m}{o-n}\div \frac{a-b}{a-c} \in\mathbb{R}$. Indeed, \begin{align*} \frac{o-m}{o-n}\div \frac{a-b}{a-c}&= \frac{0-\frac{a+b+c+abc}{2(ac+1)}}{0-\frac{a+b+c+abc}{2(ab+1)}}\div \frac{a-b}{a-c}\\ &= \frac{ab+1}{ac+1}\cdot \frac{a-c}{a-b}\\ &=\frac{\left(\frac{1}{ab}+\frac 11\right)\left( \frac 1a-\frac 1c\right)}{\left(\frac{1}{ac}+\frac 11\right)\left( \frac 1a-\frac 1b\right)} \end{align*} where the last equality follows from $$\frac{\left(\frac{1}{ab}+\frac 11\right)\left( \frac 1a-\frac 1c\right)}{\left(\frac{1}{ac}+\frac 11\right)\left( \frac 1a-\frac 1b\right)}=\frac{\left(\frac{ab+1}{ab}\right)\left( -\frac{a-c}{ac} \right)}{\left(\frac{ac+1}{ac}\right)\left( -\frac{a-b}{ab} \right)} =\frac{ab+1}{ac+1}\cdot \frac{a-c}{a-b}$$ so $\frac{o-m}{o-n}\div \frac{a-b}{a-c} \in\mathbb{R}\iff \measuredangle BAC=\measuredangle MON$ as desired.