a)$BD=BH,HI=HE\implies \angle{IEH}=\angle{HDB}\implies IEBD= cyclic\implies M$ lies on the radical axis of $(O)$ and $\triangle{ABC}$ Euler circle Similary we get $MN$ is the radical axis so $MN\perp OH$

Any nice solution for part b)? I have a solution with complex numbers and spiral similarity

mid point of EF, it is a easy problem

Comparing to the geometry problems from VMO in the last few years, this one is not that easy, though it's not very hard.

Here is a solution for part b) : Let $AS \cap RH = X$. $X$ is on the circumcircle of $\triangle ABC$. Spiral similarity centered at $R$ sends $F$ to $B$, $E$ to $C$ and $S$ to $X$. It is well known that $RX$ passes through the midpoint of $BC$ so we conclude that $RS$ passes through the midpoint of $EF$. Now let $BH \cap (O) = B_1$ and $ AH \cap BC = A_1$. Now we have $\triangle BFE \sim \triangle A_1HE$ (easy angle-chase) which immediately gives $\triangle BFE \sim \triangle DHB_1$ (because $A_1,E$ are midpoints of $HD,HB_1$). Now as $PD$ passes through midpoint of $HB_1$ and $\angle PBF= \angle PDH$ we conclude that $BP$ passes through the midpoint of $EF$. Analogously $CQ$ passes through the same point and we are finished.

Will prove that all lines pass thru the midpoint of $\overline{EF}$.By Pascal's on $BDCPQA$ we have that $EF,BP,CQ$ are concurrent. Claim:BP,CQ,A-symmedian are concurrent. From sine rule in $\triangle ADC$ we have that $\frac{CP}{AP}=\frac{\sin CDE}{\sin ADE}=\frac{EC\cdot AD}{AE\cdot CD}$ analogusly $\frac{AQ}{BQ}=\frac{FA\cdot DB}{BF\cdot AD}$ let the A-symmedian cut $\odot ABC$ in $W$ than $ABWC$ is harmonic and hence $\frac{BW}{CW}=\frac{AB}{AC}$.Multiplying everything this succumbs to ordinary Ceva in $\triangle ABC$ and three heights. Now that we known that the A-symmedian ,$\overline{EF},\overline{BP},\overline{CQ}$ are concurrent as $EF$ is anti-parallel to BC we have that $A$-symmedian passes thru it's midpoint.All that remains is to show that $RS$ passes thru the midpoint of $\overline{EF}$.$R$ is the Miquel's point of $CBEF$ and hence as $CBEF$ is cyclic we have theat $AR$ passes thru $BC\cap EF$.Projecting this onto $\odot AEF$ we have that $FHER$ is harmonic $\implies$ $RH$ is the symmedian in this triangle but as $AO,AH$ are isonals in $A$ so are $RS,RH$ in $R$ which implies the desired.

nikolapavlovic wrote: Will prove that all lines pass thru the midpoint of $\overline{EF}$.By Pascal's on $BDCPQA$ we have that $EF,BP,CQ$ are concurrent. Claim:BP,CQ,A-symmedian are concurrent. From sine rule in $\triangle ADC$ we have that $\frac{CP}{AP}=\frac{\sin CDE}{\sin ADE}=\frac{EC\cdot AD}{AE\cdot CD}$ analogusly $\frac{AQ}{BQ}=\frac{FA\cdot DB}{BF\cdot AD}$ let the A-symmedian cut $\odot ABC$ in $W$ than $ABWC$ is harmonic and hence $\frac{BW}{CW}=\frac{AB}{AC}$.Multiplying everything this succumbs to ordinary Ceva in $\triangle ABC$ and three heights. Now that we known that the A-symmedian ,$\overline{EF},\overline{BP},\overline{CQ}$ are concurrent as $EF$ is anti-parallel to BC we have that $A$-symmedian passes thru it's midpoint.All that remains is to show that $RS$ passes thru the midpoint of $\overline{EF}$.$R$ is the Miquel's point of $CBEF$ and hence as $CBEF$ is cyclic we have theat $AR$ passes thru $BC\cap EF$.Projecting this onto $\odot AEF$ we have that $FHER$ is harmonic $\implies$ $RH$ is the symmedian in this triangle but as $AO,AH$ are isonals in $A$ so are $RS,RH$ in $R$ which implies the desired. Can you explain to me what is $A$-symmedian?

It's a Symmedian coming out of vertex $A$.