VMO

a) First, applying Pascal's theorem for 6 points ($DPN$, $MEF$), we get the intersections of pairs of lines ($DE,MP$); ($DF,MN$); ($PF,NE$) collinear or $A$, $X$ and $Y$ are collinear.



Clearly, $180^{\circ }-\angle BAC=\angle BHC=\angle BKC$ so $K$ lies on $(O)$. Next, we will prove that $XY$ bisects $EF$.

Note that $MN$, $MP$ are the midlines of the triangle $ABC$ so $MN\parallel AB$, $MP\parallel AC$. Therefore, by Thales's theorem, we have $$\frac{YD}{YF}=\frac{MD}{MB},XE=\frac{MC}{MD}$$ Let $Q$ be the intersection of $XY$ and $EF$. Applying Menelaus' theorem to triangle $DEF$, we get $$\frac{QF}{QE}=\frac{XD}{XE},\frac{YF}{YD}=\frac{MB}{MD},\frac{MD}{MC}=-1$$ or $Q$ is the midpoint of $EF$. It follows that $AQ$, $AM$ are isogonal with respect to angle $\angle BAC$, so $AQ$ is the symmedian of triangle $ABC$. Therefore, $ABZC$ is a harmonic quadrilateral.

Let $J$ be the intersection of $EF$ and $BC$, then $(JD,BC)=-1$ so $K(JD,BC)=-1$ but we also have $K(ZA,BC)=-1$, which implies $KZ$ passes through $J$.

Finally, we have $JE\cdot JF=JB\cdot JC=JK\cdot JZ$ so $K,Z,E$ and $F$ are concyclic.

b) We have $\angle EBF=\angle ECH=\angle EDH$, $\angle HED=\angle HCD=\angle BEF$ because $BCEF$, $EHDC$ are cyclic quadrilaterals. Therefore, $△BEF\sim △DEH$ (a.a). Hence, $$\frac{HK}{2EH}=\frac{HD}{EH}=\frac{BF}{EF}=\frac{BF}{2F{Q}^{\prime }}$$ but we also have $\angle BFQ=\angle EHK$, then $△BFQ\sim △SLE$ (s.a.s), which implies $$\angle FBQ=\angle EKH=\angle ABS$$ or $B$, $Q$ and $S$ are collinear. Similarly, $C$, $T$ and $S$ are collinear. Therefore, $BS$, $CT$ and $XY$ are concurrent at $Q$. $\square$