Actually, we don't need condition $OH\parallel BC$ as follows $A$-Altitude of triangle $ABC$ meets its circumcircle $(O)$ again at $D$. $DC$ and $DB$ meet the parallel line from $O$ to $BC$ at $M$ and $N$, respectively. Let $E$ and $F$ be the orthogonal projection of $M$ and $N$ on the lines $CA$ and $AB$, respectively. Prove that $EF$ bisects the segments $BC$.

buratinogigle wrote: Actually, we don't need condition $OH\parallel BC$ as follows $A$-Altitude of triangle $ABC$ meets its circumcircle $(O)$ again at $D$. $DC$ and $DB$ meet the parallel line from $O$ to $BC$ at $M$ and $N$, respectively. Let $E$ and $F$ be the orthogonal projection of $M$ and $N$ on the lines $CA$ and $AB$, respectively. Prove that $EF$ bisects the segments $BC$. Oops... I didn't notice that $H$ is useless when testing this problem lol First, by angle chasing, we can show that $$ \measuredangle NAC = \measuredangle NAD + \measuredangle DAC = \measuredangle ADN + \measuredangle DBC = 90^{\circ}. $$Similarly, $\measuredangle BAM = 90^{\circ}$. Therefore, $\measuredangle NAF = \measuredangle MAE$ and it follows that $\triangle NAF \stackrel{-}{\sim} \triangle EAM$. We then have $$ \frac{\overline{AN}}{\overline{AM}} = \frac{\overline{AF}}{\overline{AE}}. $$ Since $A$ and $D$ are symmetric with respect to the line $MN$ and $BC \parallel MN$, we have $$ \frac{\overline{AN}}{\overline{AM}} = \frac{\overline{DN}}{\overline{DM}} = \frac{\overline{DB}}{\overline{DC}} $$ On the other hand, $\triangle BNF \stackrel{+}{\sim} \triangle CEM$ because $\measuredangle FBN = \measuredangle ECM$, we have $$ \frac{\overline{FB}}{\overline{EC}} = \frac{\overline{BN}}{\overline{CM}} = \frac{\overline{DB}}{\overline{DC}}. $$ Combining all results above, we have $$ \frac{\overline{AF}}{\overline{FB}} \times \frac{\overline{BM}}{\overline{MC}} \times \frac{\overline{CE}}{\overline{EA}} = -\frac{\overline{AF}}{\overline{AE}} \times \frac{\overline{CE}}{\overline{FB}} = \frac{\overline{AN}}{\overline{AM}} \times \frac{\overline{DC}}{\overline{DB}} = -1. $$By Menelaus' theorem, we know $E, F, M$ are collinear, as desired.

More general problem. Let $ABC$ be a triangle inscribed in circle $(O)$. $M$ is the midpoint of $BC$. $D$ is any point on $(O)$. $OM$ meets $(AOD)$ again at $P$. Let $Q$ and $R$ be the intersection of lines $DB$ and $DC$, respectively, with the parallel line from $P$ to $BC$. Let $F$ and $E$ be the orthogonal projections of $Q$ and $R$ on the lines $AB$ and $AC$, respectively. Prove that $E$, $F$ and $M$ are collinear. Proof. Let $(O)$ be unit circle in the complex plane. We have $$\bar a=\frac{1}{a}, \bar b=\frac{1}{b}, \bar c=\frac{1}{c}, \bar d=\frac{1}{d}, m=\frac{b+c}{2}, \bar m=\frac{1}{2b}+\frac{1}{2c}.$$Since $OM$ meets $(AOD)$ again at $P$, we get $$p=\frac{ad + bc}{a + d},\bar p=\frac{ad + bc}{abc + bcd}.$$Since $Q$ and $R$ are the intersections of lines $DB$ and $DC$, respectively, with the parallel line from $P$ to $BC$, we get $$q=\frac{-2ad^{2} + cd^{2} + abc + acd - bcd}{-d^{2} + ac - ad + cd},\bar q=\frac{-d^{2} - ab + ad + 2bc - bd}{-bd^{2} + abc - abd + bcd}$$and $$r=\frac{-2ad^{2} + bd^{2} + abc + abd - bcd}{-d^{2} + ab - ad + bd},\bar r=\frac{-d^{2} - ac + ad + 2bc - cd}{-cd^{2} + abc - acd + bcd}.$$Since $F$ and $E$ are the orthogonal projections of $Q$ and $R$ on the lines $AB$ and $AC$, respectively, we get $$e=\frac{ab + ac - 2ad + bd - cd}{2b - 2d},\bar e=\frac{ab - ac + 2bc - bd - cd}{2abc - 2acd}$$and $$f=\frac{ab + ac - 2ad - bd + cd}{2c - 2d},\bar f=\frac{-ab + ac + 2bc - bd - cd}{2abc - 2abd}.$$From these, it is not hard to check $$\frac{m-e}{m-f}+\frac{\bar m-\bar e}{\bar m-\bar f}=0.$$This means $E$, $F$ and $M$ are collinear.

buratinogigle wrote: More general problem. Let $ABC$ be a triangle inscribed in circle $(O)$. $M$ is the midpoint of $BC$. $D$ is any point on $(O)$. $OM$ meets $(AOD)$ again at $P$. Let $Q$ and $R$ be the intersection of lines $DB$ and $DC$, respectively, with the parallel line from $P$ to $BC$. Let $F$ and $E$ be the orthogonal projections of $Q$ and $R$ on the lines $AB$ and $AC$, respectively. Prove that $E$, $F$ and $M$ are collinear. Let's solve the generalization of buratinogigle where $D$ lies on arc $BC$. \[\measuredangle APO=\measuredangle ADO=90-\measuredangle DCA=90-\measuredangle RCF=\measuredangle FRC\]Similarily \[\measuredangle APO=\measuredangle ADO=\measuredangle ABD-90=\measuredangle EQB\]Hence we have $BEQ\sim CFR$. Also \[\measuredangle QPA=90-\measuredangle APO=\measuredangle QBA\]Subsequently, $A,Q,B,P$ are concyclic. Similarily $A,R,F,P$ are concyclic. \[\measuredangle BAQ=\measuredangle BPQ=\measuredangle PBC=\measuredangle BCP=\measuredangle RPC=\measuredangle RAC\]Thus, $AEQ\sim AFR$. By these similarities, we conclude that \[\frac{FC}{FA}.\frac{EA}{EB}=\frac{FC}{EB}.\frac{EA}{FA}=\frac{CR}{BQ}.\frac{EQ}{FR}=1=\frac{MB}{MC}\]By Menelaus, we get that $E,M,F$ are collinear which is the desired result.$\blacksquare$