Let $ABC$ be triangle with circumcircle $(O)$ of fixed $BC$, $AB \ne AC$ and $BC$ not a diameter. Let $I$ be the incenter of the triangle $ABC$ and $D = AI \cap BC, E = BI \cap CA, F = CI \cap AB$. The circle passing through $D$ and tangent to $OA$ cuts for second time $(O)$ at $G$ ($G \ne A$). $GE, GF$ cut $(O)$ also at $M, N$ respectively.
a) Let $H = BM \cap CN$. Prove that $AH$ goes through a fixed point.
b) Suppose $BE, CF$ cut $(O)$ also at $L, K$ respectively and $AH \cap KL = P$. On $EF$ take $Q$ for $QP = QI$. Let $J$ be a point of the circimcircle of triangle $IBC$ so that $IJ \perp IQ$. Prove that the midpoint of $IJ$ belongs to a fixed circle.
I think $O$ is supposed to be fixed as well?
Refer to diagram above, uses some well-known results.
(a) Note that the condition is equivalent to $\angle AGD = \frac{\angle B = \angle C}{2}$ (WLOG that $\angle B$ is bigger). This implies that $DG$ passes through the midpoint of arc $BC$ which contains $A$. By the Angle-Bisector Theorem, we have that $\frac{BG}{GC} = \frac{BD}{DC} = \frac{BA}{AC},$ in other words that $ABGC$ is harmonic. Now, by inversion at $F$ with power $-AF * FB$, we have that quadrilateral $ANLB$ is also harmonic. Let $H' = CN \cap AD$. Then, by projecting $ANLB$ from $C$ onto line $AD,$ we get that $(A, I; H', D) = -1.$ As $(A, I, EF \cap AI, D) = -1$ also, we have that $CN, AI,$ and $EF$ concur. Similarly, we see that $BM, AI,$ and $EF$ concur. Hence, we have that $H = BM \cap CN \cap AI \cap EF$. Hence, we have that $AH$ is just the internal angle bisector of $\angle BAC$, and so since $(O)$ and $BC$ are fixed, the desired fixed point is the midpoint of the smaller arc $BC$.
(b) Let $K', L'$ be the reflections of $I$ over $K, L,$ respectively. Then, let $S' = K'L' \cap IQ$. Observe that $IS' = 2IS = 4IQ$ by the condition. Also, it's well-known that $KL$ is the perpendicular bisector of $AI$. Therefore, we have that since $K', L' \in AX$ and $X \in EF$, $(S', I; Q, R) = -1$ by projecting $(A, I; H, D)$ from $X$ onto line $QR$. Therefore, combining with $IS' = 4IQ$, we get that $SI = IR$. Now, by using the converse of the Butterfly Theorem, we see that $I$ must be the midpoint of the chord of the circumcircle obtained by extending line $RS$. Therefore, we see that $SI \perp IO$, and so consequently the midpoint of $IJ$ lies on the circle of diameter $OT$, clearly fixed.
Here is my solution for part a
Solution
Let $S$ $\equiv$ $AI$ $\cap$ ($O$) ($S$ $\not \equiv$ $A$) then $S$ is fixed point
It's easy to see that: $AG$ is $A$ - symmedian of $\triangle$ $ABC$
So: $\dfrac{SB}{SC}$ . $\dfrac{MB}{MA}$ . $\dfrac{NA}{NB}$ = $\dfrac{EC}{EA}$ . $\dfrac{GA}{GC}$ . $\dfrac{FA}{FB}$ . $\dfrac{GB}{GA}$ = $\dfrac{EC}{EA}$ . $\dfrac{FA}{FB}$ . $\dfrac{GB}{GC}$ = $\dfrac{EC}{EA}$ . $\dfrac{FA}{FB}$ . $\dfrac{AB}{AC}$ = $\dfrac{EC}{EA}$ . $\dfrac{FA}{FB}$ . $\dfrac{DB}{DC}$ = 1 or $AS$, $BM$, $CN$ concurrent at $H$
Hence: $AH$ passes through fixed point $S$ is midpoint of $\stackrel\frown{BC}$ which not containing $A$
About part b, I have the same idea as Pathological, just note that: $E$ $\equiv$ $AC$ $\cap$ $GM$, $F$ $\equiv$ $AB$ $\cap$ $GN$, $H$ $\equiv$ $BM$ $\cap$ $CN$ then applying Pascal theorem for 6 point $A$, $M$, $N$, $G$, $C$, $B$ then: $E$, $F$, $H$ are collinear
Vietnam TST 2016 P3 wrote:
Let $ABC$ be triangle with circumcircle $(O)$ of fixed $BC$, $AB \ne AC$ and $BC$ not a diameter. Let $I$ be the incenter of the triangle $ABC$ and $D = AI \cap BC, E = BI \cap CA, F = CI \cap AB$. The circle passing through $D$ and tangent to $OA$ cuts for second time $(O)$ at $G$ ($G \ne A$). $GE, GF$ cut $(O)$ also at $M, N$ respectively.
a) Let $H = BM \cap CN$. Prove that $AH$ goes through a fixed point.
b) Suppose $BE, CF$ cut $(O)$ also at $L, K$ respectively and $AH \cap KL = P$. On $EF$ take $Q$ for $QP = QI$. Let $J$ be a point of the circimcircle of triangle $IBC$ so that $IJ \perp IQ$. Prove that the midpoint of $IJ$ belongs to a fixed circle.
Solution: Let $Y$ be the center of circle through $D$ and tangent to $OA$ at $A$. Let $M_A$ be the midpoint of arc $BC$ not containing $A$.
$$\angle YDA=\angle YAD=\angle ACM_A=\angle BDA \implies Y \in BC$$$CY$ is the $C-$symmedian WRT $\Delta AGC$ $\implies$ $AG$ is the $A-$symmedian WRT $\Delta ABC$. Apply Pascal on $CABMGN$ $\implies $ $H$ $\in$ $\overline{EF}$. Obviously, $GK,$ $BC,$ $AM_{BC}$ concur , say at $X$ (where $M_{BC}$ is the midpoint of arc $BAC$.) Let $I_B,I_C$ be $B,C-$ $\text{excenters} $ $\implies$ $EF$ is orthic axis WRT $\Delta I_BII_C$ $\implies$ $X \in EF$. Apply Pascal on $KABCNG$ and $KACBMG$ $\implies$ $H$ $\in$ $AM_A$ which is a fixed line.
For part (b), Showing, $OI \perp IQ$ suffices which can be done using harmonic bundles and Butterfly theorem (Same as Pathological)
Part $a$
By $\sqrt{bc}$ inversion $G$ is $A-$ symmedian intersection with $O$. Then it is easy to see that $EF,BC,TG$ are concurrent at $X$ (look at picture in post 1 for point names). Then by DDIT on quadrilateral $BCEF$ and point $G$ we get that $(GB,GE),(GC,GF),(GX,GA)$ are pairs of lines that involution swaps. By projecting involution from $G$ to $(O)$ we are done.
Another way for part B.
By Pascal theorem, the tangent to the circumcircle at $A$ , $KL$ , $EF$ and the line parallel to $BC$ at $I$ are concurrent at $Z$. Well known that $EF \perp OI_a$. Then, \[ -1=Z(I,P;Q,\infty_{IQ}) \overset{\text{rotate} 90^{\circ}}{=} O(T,\infty_{AT};I_a,P') \implies T \ \text{is the midpoint of} \ I_aP' \implies P'=I\]($T$ is the midarc of $BC$ not containing $A$ and $OP' \perp IQ$). Hence, $IQ \perp OI$ and the midpoint $IJ$ lies on a fixed circle with diameter $OT$
