Dear Mathlinkers, Pascal's theorem in a degenerated case work very well.. Sincerely Jean-Louis

Obviously, \(\angle BAO = \angle CAH\) Since \(\angle BAI = \angle CAI\), then \(\angle DAI = \angle HAI\) \(\implies \angle DAE = \angle DCE\) then \(\angle EAF = \angle DCE\) \(\implies\) AMNC cyclic \(\implies \angle MNE = \angle MAC = \angle BAE = \angle BLE\) thus \(MN || BC\)

We have: $\angle{MCN} = \angle{EAD} = \angle{MAN}$ Then: $A$, $C$, $M$, $N$ lie on a circle So: $\angle{ANM} = 180^o - \angle{ACM} = 180^o - 90^o = 90^o$ But: $AN$ $\perp$ $BC$ then: $MN$ $\parallel$ $BC$

This is relatively easy for a G10 problem, isn't it (considering no special theorems were used)? Btw I also got the same solution as Pikachu1729

It also works on any isogonal lines $AD, AF$. $\angle MCN = \angle MAN$ $\implies ACMN$ - cyclic. By Reim $MN \parallel DF \parallel BC$.

Complex numbers work very well too if you're too lasy to even draw a diagram. Let $a,b,c\in\mathbb{C}$ be such numbers lying on unit circle centered at $0$ such that $$A=a^2, B=b^2, C=c^2, E=-bc$$We have $$D=-a^2, F=-\frac{b^2c^2}{a^2}$$Hence $$M=\frac{CD(A+E)-AE(C+D)}{CD-AE}=\frac{b(c^2-a^2)+c(bc-a^2)}{b-c}$$and using analogous formula $$N=\frac{a^2bc(c-b)+c(b^2c^2-a^4)}{a^2(b-c)}$$Thus $$\frac{M-N}{B-C}=\frac{b (c^2 - a^2)(a^2 - b c)}{a^2(b-c)^2(b+c)}$$Clearly $$\frac{M-N}{B-C}=\overline{\left(\frac{M-N}{B-C}\right)}$$QED

Since $O$ and $H$ are isogonal conjugates with respect to $\triangle ABC,$ $$\measuredangle NCM=\measuredangle ECD=\measuredangle EAD=\measuredangle FAE=\measuredangle NAM$$and $ACMN$ is cyclic. Hence, $$\measuredangle BCE=\measuredangle BAE=\measuredangle MAC=\measuredangle MNC.$$$\square$

Applying pascal on $DCEEAF$, we have $DF\cap EE$ lies on $MN$. However , $\angle AFD=90^{\circ}$ and $AF\perp BC$ so $BC\parallel FD$. Thus, we need to prove that $EE \parallel FD$ which is equivalent to $EF=ED$. And it’s true because $E$ is the midpoint of arc $BC$ not containing $A$ and arc $BF$ = arc $DC$ due to $AH,AO$ being isogonal conjugate.