Part a). (See fig001) We will make use of a well-known lemma. Lemma. $\triangle ABC$ (assume that $AB<AC$) inscribed in a circle $\omega$. $D$ is a point taken on the internal bisector of $\angle BAC$ such that $DB=DC$. Then $D\in\omega$. Proof: Pretty exercise for reader! Now we attack the main problem. First by MIQUEL's theorem, $G\in (KCE)\cap (KBF)$. Hence simple angle-chasing shows that $GO$ bisects $\angle GAK$, then by applying the above lemma, we obtain that $GAOK$ is concyclic. Now note that $BC$, $OG$, $AK$ is the radical axes of pairs of circles $(O),(I),(GAO)$, hence the concurrency.

Any solution for part b

b) Let $S = BC \cap DG$, let $T$ be midpoint of $BC$ and let $T'$ be its inverse wrt. $\odot BOC$. By Brokard theorem $PQ$ is polar of $S$ wrt. $\odot BOC$. Since $BC$ is polar of $T'$ wrt $\odot BOC$ and $S \in BC$, from La'Hire theorem we conclude $T' \in PQ \implies \operatorname{pow}(T', \odot GPQ) = \operatorname{pow}(T', \odot ABC)$, which is a constant since $T'$ is fixed as $G$ varies. Let $X$ be second intersection of $\odot GPQ$ and $\odot BOC$. $GX$, $BC$ and $PQ$ are concurrent at radical center $R$ of $\odot ABC$, $\odot BOC$ and $\odot GPQ$. Since $GS$, $BN$ and $CM$ are concurrent in triangle $GBC$, we get $(BCSR) = -1$ $\implies$ $D(BCGR) = -1$. Let $Y=GT' \cap \odot GBC$, since $GT'$ is $G$-symmedian in $\triangle GBC$, we get $(BCGY) = -1$ $\implies$ $D(BCGY) = -1$ $\implies$ $R,D,Y$ are collinear. Finally we have $(CBXD) \overset{R}{=} (BCGY) = -1$ $\implies$ $X$ is fixed as $G$ varies $\implies$ all circles defined for different postitions of $G$ are coaxial with radical axis $XT'$.

Here is my solution for this problem Solution a) We have: $(KC; KB) \equiv (BF; BK) + (FK; FB) \equiv (GC; GA) + (AB; AC) + (EA; EB)$ $\equiv (GC; GB) + (AB; AC) \equiv (OC; OB) + (AB; AC) \equiv (AC; AB)$ (mod $\pi$) So: $K$ $\in$ $(O)$ But: $G$ $\equiv$ $(ACF)$ $\cap$ $(ABE)$ then: $G$ is Miquel point of completed quadrilateral $ABKC.EF$ Let $U$ $\equiv$ $BC$ $\cap$ $AK$ so by Brocard theorem, we have: $U$ is orthocenter of $\triangle$ $OEF$ or $OU$ $\perp$ $EF$ at $G$ Hence: $BC$, $AK$, $OG$ concurrent at $U$ b) Let $J$ be intersection of tangents at $B$, $C$ of $(I)$; $Y$ $\equiv$ $JD$ $\cap$ $(I)$ $(Y \not \equiv D)$ Applying Pascal theorem for 6 points $B$, $C$, $D$, $C$, $B$, $G$, we have: $J$ $\equiv$ $BB$ $\cap$ $CC$, $N$ $\equiv$ $CG$ $\cap$ $BD$, $M$ $\equiv$ $BG$ $\cap$ $CD$ are collinear But tangents at $B$, $C$ of $(I)$ and $YD$ concurrent at $J$ then: $BDCY$ is harmonic quadrilateral or $G(BCDY) = - 1$ Let $T$ $\equiv$ $MN$ $\cap$ $BC$ so: $G(BCDT) = - 1$ or $G$, $T$, $Y$ are collinear Then: $\overline{TP} . \overline{TQ} = \overline{TB} . \overline{TC} = \overline{TY} . \overline{TG}$ or $Y$ $\in$ $(GPQ)$ Let $X$ $\equiv$ $DY$ $\cap$ $(GPQ)$ $(X \not \equiv Y)$ so: $\overline{JX} . \overline{JY} = \overline{JP} . \overline{JQ} = JO^2 - R_{(O)}^2$ (With $R_{(O)}$ is radius of $(O)$) Hence: $X$ is fixed point Therefore: $(GPQ)$ passes through 2 fixed points $X$, $Y$

Part a) Trivially see from the Miquel Point and Brokard Theorem: First, it's easy to see that $G$ is the Miquel Point of the complete quadrilateral $ABKCEF$, we have $$ \angle BKF = \angle BGF = \angle ABC \implies K \in (O) $$Thus from Brokard Theorem, we have $OS \perp EF$ for $S=AK \cap BC$, also since $G$ is the Miquel Point of the concyclic quadrilateral $ABKC$, we have $OG \perp EF$ means $O,S,K$ are collinear, finish the proof! Part b): Some directly synthetic proof can be seen from all the above solutions. I'm now giving a non-synthetic solution using inversion which is clearer to see the fixed point! First, we can rewrite the problem as: " Given $(O)$, $B,C$ are $2$ fixed points lying on $(O)$, $A$ varies on $(O)$. Let $S$ be the midpoint of the arc $BC$ that doesn't contain $A$. Let $D$ be an arbitrarily fixed point on $(O)$. $BD \cap AC =F$; $CD \cap AB = E$; $EF \cap (S,SB) = P,Q$. Then prove that $(APQ)$ passes through $2$ fixed points when $A$ varies " Now for the new problem, take the inversion with center $A$ and arbitrary ratio gives a new problem: " Given $\triangle ABC$ with $(I)$ be the incenter. $D$ be a fixed point lying on $BC$. $(ABD) \cap AC=F$, $(ACD) \cap AB =E$. $(AEF) \cap (I)$ at $P,Q$. Then prove that $PQ$ passes through a fixed point when $A$ varies " Solution to the new problem Let $(AEF) \cap (O)$ at $S$. By applying the theorem of the radical axis for $3$ circles $(AEF),(O),(BIC)$, we see that $AS,PQ,BC$ are concurrent at $T$. We're now proving that $T$ is the fixed point We prove that $(BC,TD)=-1=A(BC,TD)=(B,C, AD \cap (O), S)=(BC,GS)$ From the spiral similarity with center $S$, we see that $\triangle SEB \sim \triangle SFC$ means $$ \frac{SB}{SC}=\frac{BE}{CF} $$Also since $AEDC$ and $AFDB$ are both concyclic, we have $BE.BA=BD.BC$ and $CF.CA=CD.CB$, means $$ \frac{SB}{SC}=\frac{BE}{CF} = \frac {BD}{CD}. \frac {AC}{AB} = \frac{GB}{GC} $$Thus $SBGC$ is a hamornic quadrilateral, which means $T$ is fixed and we are done!!!