Notice that $XC\|AF$. Let $AD\cap BC=T$. So, $$-1=(TF;BC)\overset{X}{=}(A,F;XB\cap AF,\infty_{AF})\implies BX\text{ bisects } AF$$

$\angle FAX=\angle AXC=90$ so $XC\|AF$. let $AB\cap XC=E$. Look at $\triangle EAC$. $AX$ is both the angle bisector and the perpendicular. Hence $\triangle EAC$ is equilateral with $AE=AC$. Then $X$ is the midpoint of $EC$. And since we have proved that$XC\|AF$, the result follows from homothety at $B$.

Let $N$ be the midpoint of segment $AC$ and let $R$ be the second intersection of $BX$ and $\omega$. Also let $L$ be the intersection of $BX$ and $AF$. It's clear that $AF$ is parallel to $XC$. $\angle RXN = \angle RXC - \angle NXC = \angle ALX - \angle NXC = \angle ALX - \angle LAC = \angle ALX - \angle LAB = \angle ABR = \angle NCR$ implying that $RXCN$ is cyclic. Thus, $\angle LAN = \angle NCX = \angle NRL$ implying that $ARLN$ is cyclic. By the last one, $\angle LNA = \angle LRA = \angle LCA$ implying that $LN$ in parallel $BC$. Since N is the midpoint of segment $AC$ we have that $L$ is the midpoint of $AF$.

Let A' be qa point on AB sush that BA=BA' and we now that XA=XA'.Now from homothety its enougth to prove that B'-M(M is the point of intersection of BC and the angle bisector) - X' are collinear.

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amar_04 wrote: Notice that $XC\|AF$. Let $AD\cap BC=T$. So, $$-1=(TF;BC)\overset{X}{=}(A,F;XB\cap AF,\infty_{AB})\implies BX\text{ bisects } AF$$ i think you should define BCnAC=Y and CQnAB=Z and prove that T-Z-Y collinear then your solution follow.Correct me if i am wrong

Since $AF$ passes through the midpoint of arc $BC$, i.e. the antipode of $D$, we know $AD \perp AF$. Now, $XC \parallel AF$ follows easily. Let $AD \cap BC = E$. Since $\angle EAF = 90^{\circ}$ and $\angle BAF = \angle FAC$, we conclude $(E, F; B, C) = -1$ by properties of Apollonian Circles. To finish, observe $$-1 = (E, F; B, C) \overset{X}{=} (A, F; M, P_\infty)$$as desired. $\blacksquare$ Remark: My GeoGebra diagram is linked here!

$$K=AF\cap BX, E=AD \cap BC, (A, F; K, P_{\infty})\stackrel{X}{=}(E, F; B, C)\stackrel{\text{Lemma~9.18}}{=}-1\implies AK=KF.\blacksquare$$

One-liner: $$-1=(\overline{XA} \cap \overline{BC}, F; BC) \stackrel X= (AF;\overline{BX} \cap \overline{AF}, \infty_{AF}).$$

D lies on the perpendicular bisector of BC $\Rightarrow$ AD is external bisector of $\angle CAB$ $\Rightarrow$ $\angle FAD = 90^{\circ}$ $\Rightarrow$ $\angle XAF = \angle CXA = 90^{\circ}$ $\Rightarrow$ $XC \parallel AF$. Denote $AD \cap BC = E$ and $AF \cap BX = K$, $\angle EAF = 90^{\circ}$, $\angle CAF = \angle FAB$ $\Rightarrow$ from lemma 9.18 (E, F; B, C) = -1. Projecting through X we get: (E, F; B, C) = (A, F, K, $P_\infty$) = -1. From (A, F, K, $P_\infty$) = -1 we get that K is midpoint of AF $\Rightarrow$ BX bisects AF and we are ready.

Suppose that $BY \perp AD$ at $Y$ and $M \equiv BX \cap CY$. It's easy to prove that $\triangle ABY \sim \triangle ACX$. Then $\dfrac{MY}{MC} = \dfrac{BY}{CX} = \dfrac{AY}{AX}$. So $AM \parallel CX$ or $A, M, F$ are collinear. But $\dfrac{AM}{CX} = \dfrac{YA}{YX} = \dfrac{BD}{BC} = \dfrac{MF}{CX}$ then $AM = MF$ or $M$ is midpoint of $AF$

What's the point of defining $D$? Let $Y$ be the foot from $B$ to the external bisector. Then, $YBCX$ is a trapezoid, and due to similarity ratios, $\overline{AF}$ passes through the intersection of diagonals $\overline{CY}$ and $\overline{BX}$. So, by well known trapezoid properties, the two diagonals $\overline{BX}$ and $\overline{CY}$ bisect $\overline{AF}$.

Let $\overline{BX} \cap \overline{AF}=G$, and $\overline{AX} \cap \overline{BC}=E$. Note that by right angles and bisectors $-1=(BC; EF) \stackrel{X}= (G \infty ; AF)$ thus $G$ is the midpoint of $\overline{AF}$ and we are finished.

Let $AD$ meet $BC$ at $L$, $BX$ intersect $AF$ at $N$ and $P_\infty$ be the point at infinity along $AF$. $$\implies (A,F;N,P_\infty) \stackrel{X}{=} (L,F; B,C)=-1$$and hence $N$ is midpoint of $AF$.

Let $Y$ be the foot from $B$ to the external bisector. Let $M=AF\cap BX$, and $P_\infty$ the point at infinity on $AF$, and $E=AD\cap BC$. Note that \[(A,F;M,P_\infty) \stackrel{X}{=} (A,D; X,Y)\stackrel{P_\infty}{=} (F,D,C,B)=-1\]from right angles + bisector lemma.

Notice that $AF \parallel CX$. Let $AX \cap BC = E$. Then we have $$-1 = (EF;BC) =^X (AF;PP_{\infty})$$where $P = AF \cap BX, P_{\infty} = CX \cap AF$. So we're done.

Let $E$ be the midpoint of arc $BC$ and $M = AF \cap BX$. Since $AF \perp AD \perp CX$, we get \[(AF;M\infty) \overset{X}{=} (XA \cap BC, F; B,C) \overset{A}{=} (DE;BC) = -1. \quad \blacksquare\]