VMO

a) Let $I$, $X$ and $Y$ be the midpoints of $BC$, $BE$ and $CF$, respectively. Let $IX$, $IY$ intersect $AC$, $AB$ at $S$, $T$ respectively. Since $IB$ is tangent to $(ABG)$, we have $$I{B}^2=IG\cdot IA=I{C}^2,$$ so $IC$ is tangent to $(AGC)$ as well. Since $I$, $Y$ are midpoints of $BC$, $CF$, we have $IY$ is parallel to $BF$, then $\angle IYG=\angle BFG=\angle BAG$, and thus $A,T,Y$ and $G$ are concyclic, so $IY\cdot IT=IG\cdot IA$.

Similarly, $$IS\cdot IX=IG\cdot IA=IY\cdot IT,$$ implying that $A,T,S,Y,X$ and $G$ lie on the same circle.



Hence, $\angle AXY=\angle AGY=\angle IGC=\angle ACI=\angle ABF=\angle ABM$. Therefore, $A,M,B$ and $X$ lie on the same circle. Similarly, $A,Y,C$ and $N$ lie on the same circle. Thus, we have $\angle AMX=\angle GBD$, and $\angle NAY=\angle GCD$. So we have $$\begin{array}{rl}\angle MAN& =\angle AMX+\angle NAY-\angle YAX\\ & =\angle GBD+\angle GCD+\angle BGC-180^{\circ }\\ & =180^{\circ }-\angle BDC,\end{array}$$ implying that $A,D,M$ and $N$ lie on the same circle.

b) For a triangle $ABC$, define the transformation namely ${\gamma }_{A,△ABC}=T\circ I$, which is the union of the inversion $I$ of center $A$, power $AB\cdot AC$, and the reflection $T$ through the angle bisector of $\angle ABC$. First, let prove the following lemma

Lemma. Let $ABC$ be a triangle and $(I)$ be a circle through $B,C$. Assume that ${\gamma }_{A,△ABC}$ transforms $(I)$ into $(J)$, then $AI,AJ$ are isogonal with respect to $\angle ABC$.



Proof. We have the inversion with center $A$ and power $AB\cdot AC$ transforms $(I)$ into $(J^{\prime })$, and $A,I,J$ are collinear due to the property of the inversion. Then after the reflection through the bisector of $\angle BAC$, $J^{\prime }$ transforms to $J$. This implies that $AI,AJ$ are isogonal with respect to angle $BAC$. $\square$

Back to our main problem, let $O_a,O_b$ and $O_c$ be the centers of $(HEF)$, $(HKF)$ and $(HKE)$. Note that $O_a,O_b,O_c$ are the mutual intersections of the perpendicular bisectors of $HK,HE,HF$, so the perpendicular bisectors of $HK,HE$ and $HF$ are precisely the lines $O_b{O}_c,O_c{O}_a$ and $O_a{O}_b$. By Desargue's theorem, $R,P$ and $Q$ are collinear if and only if $A{O}_a,B{O}_b$ and $C{O}_c$ are concurrent. So we will show that indeed $A{O}_a,B{O}_b$ and $C{O}_c$ concur.



Let $W$ be the center of the circle $\omega \equiv (KEF)$. Now let's consider the transformation ${\gamma }_{A,△ABC}$. Since $\angle BAD=\angle KAC$ and $\angle ABD=180^{\circ }-\angle C=\angle AGC$, we have $△ABD\sim △AGC$, and thus $$AB\cdot AC=AG\cdot AD,$$ and $AG,AD$ are isogonal with respect to $\angle BAC$. So ${\gamma }_{A,△ABC}$ turns $D$ to $G$ and vice versa, and hence it transforms $(BHC)$ to $(BKC)$ and vice versa. Since $H\in (BHC)$, $K\in (BKC)$ and $AH,AK$ are isogonal with respect to $\angle BAC$, the transformation will transform $H$ to $K$ and vice versa. Now, we have $$\angle ABF-\angle ACB=\angle KGC=\angle AEC,$$ and similarly $\angle ACE=\angle AFB$, then $△AFB\sim △ACE$, thus $$AE\cdot AF=AB\cdot AC.$$ Also, since $\angle FAB=\angle BAC=\angle CAE$, $AE$, $AF$ are isogonal with respect to $\angle BAC$. Hence ${\gamma }_{A,\Delta ABC}$ turns $E,F$ to each other and hence it turns $(HEF)$ and $(KEF)$ to each other. By the lemma, we have $A{O}_a$ and $AW$ are isogonal with respect to angle $BAC$.

Using this similar argument, we can show that ${\gamma }_{B,\Delta ABC}$ turns $H,E$ to each other, by noting that $△AHB\sim △ECB$ so $BA\cdot BC=BH\cdot BE$, and $\angle ABH=\angle BAG=\angle GBC$. It also turns $F,G$ to each other since $$\angle ABF=\angle ACB=\angle FBC$$ and $△ABF\sim △KBC$, so $BK\cdot BF=BA\cdot BC$. Thus, this transformation turns $(HFK)$ and $(EFK)$ to each other, and by the lemma, $B{O}_b,BW$ are isogonal with respect to $\angle ABC$. Similarly, we can show that $C{O}_c,CW$ are isogonal with respect to $\angle ACB$. Hence, $A{O}_a,B{O}_b$ and $C{O}_c$ will concur on the isogonal conjugate of $W$ with respect to triangle $ABC$. The proof is completed. $\square$