$\measuredangle AXY = \measuredangle ACB = \measuredangle YAP = \measuredangle APY$ so $AYXP$ is cyclic. Let $K=YQ \cap XF$. $\measuredangle QKX = \measuredangle QKF =\measuredangle EFK = \measuredangle ECX = \measuredangle ABX = \measuredangle QBX$ so $BXQK$ is cyclic. $\measuredangle XKY = \measuredangle XKQ = \measuredangle XBQ = \measuredangle XBA = \measuredangle XAY$ so $AYXPK$ is cyclic, finish off with $$FP \cdot FA = FX \cdot FK = FQ \cdot FB ~ \blacksquare$$ (Note that in this proof we didn't use the fact that $K \in AC$.)

Spend 3 hours on this problem!, and was it worth it?, every second was! Claim 1: Let $YQ \cap AC=G$ then $AYXPG$ is cyclic. Proof: By angle chase: $$\angle AGY=\angle ACB=\angle YAP=\angle YPA \implies AYPG \; \text{cyclic}$$ By more angle chase: $$\angle YXA=\angle ACB=\angle AGY \implies AYXG \; \text{cyclic}$$ Claim 2: Let $(BXF) \cap (GQF)=J$ then $J \in (AYXPG)$ (which is posible thanks to Claim 1) Proof: By angle chase: $$\angle XJG=\angle XJF+\angle FJG=\angle XBA+\angle YQB=\angle XBA+\angle ABC=\angle XBC=180-\angle XAG$$ Which completes the proof Claim 3: $X,F,G$ are colinear. Proof: By angle chase: $$\angle QGX=\angle YAX=\angle QBX \implies XBGQ \; \text{cyclic}$$ Now we do even more angle chase using Claim 2 $$\angle XJF=\angle XBF=\angle FGQ=\angle FJQ$$ Now by even more angle chase using the tangent line $AY$ $$\angle FJB=\angle YXG=180-\angle YAC=\angle ABC=\angle AQG=\angle FJG$$ By substracting the results on the last 2 angle chases we have $\angle XJB=\angle QJG$ and by angle chase $$\angle XFB=\angle XJB=\angle QJG=\angle QFG \implies X,F,G \; \text{colinear}$$ Main solution: Since by Claim 3 we have $X,F,G$ colinear and $XBGQ$ cyclic and by Claim 1 we have $AXPG$ cyclic and by definition $F$ is midpoint of $AB$ then by PoP we have that: $$AF \cdot FP=XF \cdot FG=BF \cdot FQ \implies FP=FQ \implies F \; \text{midpoint of} \; PQ$$ Thus we are done Note: I claim that $JG \cap BC=K$ is the projection from $A$ to $BC$, can anyone prove it?

Here's my solution: Let $Z = YQ \cup XF$. Claim. $X$, $B$, $Z$, $Q$ are concyclic.

Claim. $Y$, $X$, $P$, $Z$, $A$ are concyclic.

Finally, by PoP we have $$FP \cdot FA = XF \cdot FZ = BF \cdot FQ \Longrightarrow FP=FQ$$ as desired. $\blacksquare$

[asy][asy] import olympiad;import geometry; size(8cm);defaultpen(fontsize(10pt)); pair A,B,C,E,F,X,Y,D,P,Q,Y,A1,S,T; A=dir(110);B=dir(205);C=dir(335);E=midpoint(A--C);F=midpoint(A--B);X=intersectionpoints(circumcircle(C,E,F),circumcircle(A,B,C))[1]; D=extension(X,F,A,C);P=intersectionpoints(circumcircle(A,D,X),A--B)[1];Y=intersectionpoints(B--100X-99B,circumcircle(A,D,X))[1];Q=extension(A,B,D,Y);A1=intersectionpoints(circumcircle(A,B,C),X--100D-99X)[0];T=A*B/A1;S=intersectionpoints(circumcircle(A,D,X),A--T)[1]; draw(A--B--C--cycle,heavyred+0.5); draw(circumcircle(A,B,C),royalblue); draw(circumcircle(A,D,X),royalblue); draw(arc(circumcenter(C,E,F),C,X),royalblue); draw(A1--X,lightred+0.5); draw(B--Y--A,lightred+0.5); draw(Y--D,lightred+0.5); draw(A--T,lightred+0.5); draw(T--A1,darkred); draw(S--D,darkred); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$X$",X,dir(X)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$Y$",Y,dir(Y)); dot("$D$",D,dir(D)); dot("$A'$",A1,dir(A1)); dot("$S$",S,dir(S)); dot("$T$",T,dir(T)); [/asy][/asy] Observe that $P$ lies on $(AYX)$ as $\measuredangle APY=\measuredangle YAP=\measuredangle ACB=\measuredangle AXB=\measuredangle AXY$. Claim: Let $D=\overline{XF}\cap \overline{AC}$ and $A'=\overline{XF}\cap (ABC)$. Then, $D$ lies on $(AXYP)$ and $\overline{DY}\parallel \overline{BC}\parallel\overline{AA'}$. Proof. By Reim's theorem, we have $\overline{AA'}\parallel \overline{EF}\parallel \overline{BC}$. Therefore, $$\begin{align*} \measuredangle{YAD}&=\measuredangle{YAB}+\measuredangle{BAC}\\&=\measuredangle{AEF}+\measuredangle{BXC}\\&=\measuredangle{CEF}+\measuredangle{BXC}\\&=\measuredangle{CXF}+\measuredangle{BXC}\\&=\measuredangle{BXD}=\measuredangle{YXD}, \end{align*}$$ which means that $D$ lies on $(AXYP)$. Moreover, $\measuredangle{ADY}=\measuredangle{AXY}=\measuredangle{AXB}=\measuredangle{ACB}$, which yields that $\overline{DY}\parallel \overline{BC}$, i.e. $D$ lies on $\overline{QY}$. Introduce $S,T$ on $(AXYPD)$ and $(ABC)$ such that $\overline{AB}\parallel \overline{DS}\parallel \overline{A'T}$. By Reim's, $A,S,T$ are collinear. Hence, $$\begin{align*} -1=(A,B;F,P_\infty)\overset{A'}{=}(A,B,X,T)\overset{A}{=}(Y,P,X,S)\overset{D}{=}(Q,P,F,P_\infty). \end{align*}$$

We note that $\angle XPA = \angle XAP = \angle ACB = \angle YXA,$ so quadrilateral $AYXP$ is cyclic. Let $XY$ meet $AC$ at $I$. We now have $\angle YIA = \angle ACB = YXA$, so $I$ lies on $(AYXP)$. Now, by angle chasing, we have $$\angle BXF = \angle BXC + \angle CXF = \angle BAC + 180^{\circ} - \angle FEC = \angle BAC + \angle ACB = \angle BAC + \angle YAB = \angle YAI = \angle BXI,$$ hence $X, F, I$ are collinear. On the other hand, from the above angle chasing, $\angle BXI = 180^{\circ} - \angle ABC = \angle BQI,$ hence $BXQI$ is a cyclic quadrilateral. Now, $\overline{FP} \cdot \overline{FA} = \overline{FX} \cdot \overline{FI} = \overline{FQ} \cdot \overline{FB}$ implies $F$ is the midpoint of $PQ$, as desired.

I did it using barycentric bash. Our aim is to show that $\overline{AP} = \overline{BQ}$, as this is equivalent to $F$ being the midpoint of $PQ$. Firstly, let's define $A=(1, 0, 0), B=(0, 1, 0), C=(0, 0, 1), \overline{BC}=a, \overline{AC}=b, \overline{AB}=c$. As $E$ is the midpoint of $AC$, it follows that $E=\left(\frac12, 0, \frac12\right)$ and similary we see that $F=\left(\frac12, \frac12, 0\right)$. Then, we can calculate the equation of $(CEF)$, which gives $$-a^2yz-b^2xz-c^2xy+\frac{b^2}2x+\frac{c^2-b^2}2y = 0$$ The equation of $\Gamma$ is well known and it's $$-a^2yz-b^2xz-c^2xy = 0$$ Putting both equations together gives $$\frac{b^2}2x+\frac{c^2-b^2}2y = 0$$ So, $X$ is of the form $(b^2-c^2, b^2, t)$ for some $t\in\mathbb{R}$. Replacing this in the $\Gamma$ equation, it gives $$-a^2b^2t - b^2\left(b^2-c^2\right)t - c^2b^2\left(b^2-c^2\right) = 0$$ $$\Leftrightarrow t = -\frac{c^2\left(b^2-c^2\right)}{a^2+b^2-c^2}$$ which means that $X = \left(b^2-c^2, b^2, -\frac{c^2\left(b^2-c^2\right)}{a^2+b^2-c^2}\right)$. Then, we see that $BX = \left(b^2-c^2, t, -\frac{c^2\left(b^2-c^2\right)}{a^2+b^2-c^2}\right)$, for $t\in\mathbb{R}$. It is a well know fact that the tangent of $A$ wrt $\Gamma$ is given by $\left(t, b^2, -c^2\right)$, for $t\in\mathbb{R}$. Then, as $Y$ is the intersection of $BX$ and that tangent, we see that the tangent can be written as $\left(t, \frac{b^2\left(b^2-c^2\right)}{a^2+b^2-c^2}, -\frac{c^2\left(b^2-c^2\right)}{a^2+b^2-c^2}\right)$, and that $Y = \left(b^2-c^2, \frac{b^2\left(b^2-c^2\right)}{a^2+b^2-c^2}, -\frac{c^2\left(b^2-c^2\right)}{a^2+b^2-c^2}\right)$. In order to put $Y$ homogeneous, we sum $$b^2-c^2 + \frac{b^2\left(b^2-c^2\right)}{a^2+b^2-c^2} -\frac{c^2\left(b^2-c^2\right)}{a^2+b^2-c^2} = \frac{(b-c)(b+c)\left(a^2+2b^2-2c^2\right)}{a^2+b^2-c^2}$$ and by dividing all coordinates by that we get $Y = \left(\frac{a^2+b^2-c^2}{a^2+2b^2-2c^2}, \frac{b^2}{a^2+2b^2-2c^2}, -\frac{c^2}{a^2+2b^2-2c^2}\right)$. As $BC \parallel QY$, it follows that $Q_x = Y_x$. As $Q_z = 0$ and $Q_x+Q_y+Q_z = 1$, we check that $Q = \left(\frac{a^2+b^2-c^2}{a^2+2b^2-2c^2}, \frac{b^2-c^2}{a^2+2b^2-2c^2}, 0\right)$. Therefore, $\overline{BQ} = Q_xc = \frac{c\left(a^2+b^2-c^2\right)}{a^2+2b^2-2c^2}$. Then, we calculate $\overrightarrow{AY} = \left(\frac{b^2-c^2}{a^2+2b^2-2c^2}, \frac{b^2}{a^2+2b^2-2c^2}, -\frac{c^2}{a^2+2b^2-2c^2}\right)$, which leads to $$\overline{AY}^2 = \frac{a^2b^2c^2}{\left(a^2+2b^2-2c^2\right)^2} + \frac{b^2c^2\left(b^2-c^2\right)}{a^2+2b^2-2c^2} - \frac{b^2c^2\left(b^2-c^2\right)}{a^2+2b^2-2c^2} = \frac{a^2b^2c^2}{\left(a^2+2b^2-2c^2\right)^2}$$ As $\overline{AC}>\overline{AB}$, $b>c$, so $a^2+2b^2-2c^2 > 0$ and $\overline{AY} = \frac{abc}{a^2+2b^2-2c^2}$. As $\overline{YP} = \overline{YA}$, it follows that $\overline{PA} = \overline{YA}\times 2\cos{\angle YAP}$. We can check that $\angle YAP= \angle ACB$ and by the law of cosines in $[ABC]$, we see that $2\cos{\angle ACB} = \frac{a^2+b^2-c^2}{ab}$. Then, $\overline{PA} = \frac{abc}{a^2+2b^2-2c^2}\times\frac{a^2+b^2-c^2}{ab} = \frac{c\left(a^2+b^2-c^2\right)}{a^2+2b^2-2c^2}$. So, we get that $\overline{AP} = \overline{BQ}$. $\blacksquare$

nixon0630 wrote: Here's my solution: Let $Z = AC \cup YQ$. Claim. $Y$, $X$, $P$, $Z$, $A$ are concyclic.

Claim. $X$, $B$, $Z$, $Q$ are concyclic.

Finally, by PoP we have $$FP \cdot FA = XF \cdot FZ = BF \cdot FQ \Longrightarrow FP=FQ$$ as desired. $\blacksquare$ You didn't prove that X,F,Z are colinear

We use directed angles$\mod180^\circ$. Let $S$ be the intersection of $\overline{YQ}$ with $\overline{AC}$. Note that $$\measuredangle APY = \measuredangle YAP = \measuredangle YAB = \measuredangle ACB = \measuredangle AXB = \measuredangle AXY,$$ so $AYXP$ is cyclic. Also, since $\measuredangle QPY=\measuredangle APY=\measuredangle ACB$ and $\measuredangle YQP=\measuredangle CBA$, we have $\measuredangle PYS=\measuredangle PYQ=\measuredangle BAC=\measuredangle PAS$, so $S\in(AYXP)$. Now, $$\measuredangle BXP = \measuredangle YXP = \measuredangle YAP = \measuredangle ACB = \measuredangle AEF = \measuredangle CEF = \measuredangle CXF,$$ which gives $\overline{XP}$ and $\overline{XC}$ are isogonal with respec to $\triangle BXF$. Due to this, $$\measuredangle PXF = \measuredangle BXC = \measuredangle BAC = \measuredangle PAS = \measuredangle PXS,$$ giving that $X,F,S$ are collinear. Now, let $R$ be the point in $(AYXPS)$ such that $\overline{RS}\parallel\overline{AB}$ and $T$ be the second interseciton of $\overline{AR}$ with $\Gamma$. Notice that $$\measuredangle BXT = \measuredangle BAT = \measuredangle PAR = \measuredangle SPA = \measuredangle SXA = \measuredangle FXA,$$ giving $\overline{XT}$ and $\overline{XF}$ are isogonal with respect to $\triangle XAB$. Thus, $\overline{XT}$ is the $X-$symmedian of $\triangle XAB$ and $AXBT$ is harmonic. Let $\infty$ be the point at infinity of $\overline{AB}$. Then, $$-1=(AB;XT)\stackrel A= (YP;XR)\stackrel S= (QP;F\infty),$$ implying that $F$ is the midpoint of $\overline{PQ}$. $\blacksquare$

Let $T = \overline{AA} \cap \overline{BC}$. [asy][asy] import graph; size(12cm); pen zzttqq = rgb(0.6,0.2,0.); pen yqqqqq = rgb(0.50196,0.,0.); pen qqwuqq = rgb(0.,0.39215,0.); pen yqqqyq = rgb(0.50196,0.,0.50196); pen xdxdff = rgb(0.49019,0.49019,1.); pen cqcqcq = rgb(0.75294,0.75294,0.75294); pair A = (-4.55630,4.74594), B = (-6.,-1.), C = (2.,-1.), F = (-5.27815,1.87297), O = (-2.,1.04932), X = (-6.48864,1.27690), Y = (-6.88720,3.13406), Q = (-4.96129,3.13406), P = (-5.59500,0.61187), T = (-12.86540,-1.), Z = (-1.49456,-3.41657); draw(A--B--C--cycle, linewidth(0.6) + zzttqq); draw(A--B, linewidth(0.6) + zzttqq); draw(B--C, linewidth(0.6) + zzttqq); draw(C--A, linewidth(0.6) + zzttqq); draw(circle((-2.,1.04932), 4.49441), linewidth(0.6) + yqqqqq); draw(Y--Q, linewidth(0.6) + qqwuqq); draw(B--Y, linewidth(0.6) + qqwuqq); draw(C--X, linewidth(0.6) + yqqqyq); draw(Y--P, linewidth(0.6) + qqwuqq); draw(T--B, linewidth(0.6) + yqqqqq); draw(X--(0.55630,4.74594), linewidth(0.6) + xdxdff); draw(A--T, linewidth(0.6) + red); draw((0.55630,4.74594)--Z, linewidth(0.6) + xdxdff); draw((-1.27815,1.87297)--F, linewidth(0.6) + yqqqyq); dot("$A$", A, dir((5.364, 14.027))); dot("$B$", B, dir((-24.637, -60.663))); dot("$C$", C, dir((16.031, -39.399))); dot("$E$", (-1.27815,1.87297), dir((5.027, 10.705))); dot("$F$", F, dir((26.514, -36.078))); dot("$O$", O, dir((6.330, 10.846))); dot("$X$", X, dir((-40.985, -38.847))); dot("$Y$", Y, dir((-49.329, 9.349))); dot("$Q$", Q, dir((17.510, -16.168))); dot("$P$", P, dir((19.922, -17.709))); dot("$T$", T, dir((-44.089, -57.828))); dot("$Z$", Z, dir((-8.773, -64.261))); dot("$A'$", (0.55630,4.74594), dir((5.876, 14.027))); [/asy][/asy] Claim: [Main claim] $$\frac{AX}{BX} = \frac{AT}{AB}.$$ Proof. Let $A'$ denote the point such that $ABCA'$ is an isosceles trapezoid with $AA' \parallel BC$. Note that $A'$ lies on line $FX$, because $\measuredangle CXF = \measuredangle CEF = \measuredangle AEF = \measuredangle ACB$. Now let $Z$ be the harmonic conjugate of $X$ with respect to $AB$. Projecting $AXBZ$ through $A'$ onto line $AB$ we find $\overline{A'Z} \parallel \overline{AB}$. Finally from $\triangle BAT \sim \triangle A'AC$, we get $$\frac{AX}{XB} = \frac{AZ}{ZB} = \frac{BA'}{AA'} = \frac{AC}{AA'} = \frac{AT}{AB}. \qquad\blacksquare$$ Since $$\measuredangle APY = \measuredangle YAP = \measuredangle YAB = \measuredangle ACB = \measuredangle AXB = \measuredangle AXY$$ we get $YXPA$ is cyclic and hence $$BP \cdot BA = BX \cdot BY = BX \cdot \frac{AB \cdot AY}{AX}$$ with the last equality due to $\triangle BYA \sim \triangle AYX$. Meanwhile, $$AQ = \frac{AY}{AT} \cdot AF = \frac{AY}{AT} \cdot AB.$$ Combining with the main claim, we have $AQ = BP$ as needed.

Let $\angle XBA=t$. $A' \in (ABC)$ and $AA' \parallel BC$. $A',F,P$ are collinear because $\angle A'XC=\angle A'AC=\angle AEF=\angle FXC$. Applying sin theorem in triangle $AYQ$ and $AXB$ implies (1), $BYP$ and $AA'B$ implies (2): (1) $$BP=YP \frac{\sin(C-t)}{sin(t)}=YP \frac{XB}{XA}$$ (2) $$AQ=AY \frac{\sin(B-C)}{\sin(C)}=AY \frac{A'A}{A'B}$$ We are done because $\frac{XB}{XA}=\frac{A'A}{A'B}$ from ratio lemma.

Here's another approach using trigonometry in the inversed figure. We'll prove that $\overline{AP} + \overline{AQ} = \overline{AB}$. Firstly, let's invert around $A$ with radius $1$. Let $R'$ be the inverted point of $R$. As $E$ and $F$ are the midpoints of $AC$ and $AB$, it follows that $C'$ and $B'$ are the midpoints of $AE'$ and $AF'$, respectively. The circle $(CEF)$ inverts to the circle $(C'E'F')$ and $\Gamma$ inverts to $B'C'$, so $X'$ is the intersection of $(C'E'F')$ and $B'C'$. The line $BX$ inverts to the circle $(AB'X')$ and the tangent to $\Gamma$ through $A$ inverts to the line through $A$ parallel to $B'C'$. As $\overline{YP} = \overline{YA}$, $\angle APY = \angle YAP$, thus, in the inverted picture, it stays $\angle P'Y'A = \angle Y'AP'$. As $YQ \parallel BC$, it follows that $\angle AQY = \pi - \angle YQB = \pi - \angle CBA$, therefore, in the inverted figure, $\angle Q'Y'A = \pi - \angle AC'B'$. Finally, we check that the condition $\overline{AP} + \overline{AQ} = \overline{AB}$ becomes $\frac1{\overline{AP'}} + \frac1{\overline{AQ'}} = \frac1{\overline{AB'}}$. Let's define $\angle B'AC' = \alpha$, $\angle C'B'A = \beta$ and $\angle AC'B' = \gamma$. Let's suppose wlog that the circunradius of $[AB'C']$ is $\frac12$. Thus, $\overline{AB'} = \overline{B'F'} = \sin{\gamma}$. Now, let's see that $\angle P'Y'A = \angle Y'AP' = \pi - \angle AB'X' = \angle X'Y'A$, so, $Y'$, $X'$ and $P'$ are collinears. Then, we check that $\angle F'X'B' = \pi - \angle C'E'F' = \pi - \gamma$. So, as $\angle X'B'F' = \beta$, we see that $\angle B'F'X' = \pi - (\pi - \gamma) - \beta = \gamma - \beta$. Using the law of sines in $[X'F'B']$, $\overline{X'B'} = \frac{\overline{B'F'}\sin{\angle B'F'X'}}{\sin{F'X'B'}} = \sin{\gamma - \beta}$. As $[AY'X'B']$ is an isosceles trapezium, it follows that $$\overline{AY'} = \overline{B'X'} + 2\overline{AB'}\cos{Y'AB'} = \sin{\gamma - \beta} + 2\sin{\gamma}\cos{\beta}$$ As $[AY'P']$ is isosceles, we can calculate $$\overline{AP'} = \frac{\overline{AY'}}{\cos{Y'AP'}} = \frac{\sin{\gamma - \beta} + 2\sin{\gamma}\cos{\beta}}{2\cos{\beta}}$$ Now, we see that $$\angle AQ'Y' = \pi - \angle Q'Y'A - \angle Y'AQ' = \pi - (\pi - \gamma) - \beta = \gamma - \beta$$ So, by the law of sines in $[AY'Q']$, $$\overline{AQ'} = \frac{\overline{AY'}\sin{\angle Q'Y'A}}{\sin{\angle AQ'Y'}} = \frac{(\sin{\gamma - \beta} + 2\sin{\gamma}\cos{\beta})\sin{\gamma}}{\sin{\gamma - \beta}}$$ Now, we just replace in $\frac1{\overline{AB'}} = \frac1{\overline{AQ'}} + \frac1{\overline{AP'}}$, giving $$\frac1{\sin{\gamma}} = \frac{\sin{\gamma - \beta}}{(\sin{\gamma - \beta} + 2\sin{\gamma}\cos{\beta})\sin{\gamma}} + \frac{2\cos{\beta}}{\sin{\gamma - \beta} + 2\sin{\gamma}\cos{\beta}}$$ $$\Leftrightarrow \sin{\gamma - \beta} + 2\sin{\gamma}\cos{\beta} = \sin{\gamma - \beta} + 2\sin{\gamma}\cos{\beta}$$ which is true, ending the proof. $\blacksquare$

Cute

Nice problem Let $XF$ meet $YQ$ at $S$. Claim1 : $BXQS$ is cyclic. Proof : $\angle XBQ = \angle XBA = \angle XCE = \angle EFS = \angle FSQ = \angle XSQ$. Claim2 : $AYXPS$ is cyclic. Proof : $\angle AXY = \angle ACB = \angle YAP = \angle APY$ so $AYXP$ is cyclic. $YSX = \angle SFE = \angle ECX = \angle ACX = \angle YAX$ so $AYXS$ is cyclic. Now we have $PF.FA = FX.FS = QF.FB$ so $\frac{FP}{FQ} = \frac{FB}{FA} = 1$ so $FP = FQ$. we're Done.

We can also prove $S$ lies on $AC$. Let $YQ$ meet $AC$ at $S'$ we have $\angle AS'Y = \angle ACB = \angle YAP = \angle APY$ so $AYPS'$ is cyclic but we had $AYXPS$ is cyclic so $AYXPSS'$ is cyclic but then the circle meets line $YQ$ at three different points $Y,S,S'$, so $S'$ is $S$.

Part 1: Extend $FX$ to meet $\Gamma$ again at $A'$. Then $\angle A'XC \ (FXC)=\angle FEA = \angle C$. Hence arcs $AB$ and $A'C$ are equal so $AA'CB$ is an isosceles trapezium. Part 2: Consider Pascal's Theorem on $AABBXA'$. We will have $AA\cap BX = Y$, $BB\cap AA'=Z$, and $AB\cap A'X=F$ are collinear. Part 3: Note that $YP\parallel ZB$ (since $\angle YPA=\angle YAP=\angle C=\angle ZBA$), $YQ\parallel ZA$, and $PQ$ lies on $AB$. Hence $\triangle ZBA\sim\triangle YPQ$. Since $ZYF$ is the $Z-$median of $\triangle ZAB$, it is also the $Y-$median of $\triangle YPQ$. Hence $FP=FQ$.

Let $A'$ be such that $AA'BC$ is an isosceles trapezoid, and let $Z=XA'\cap AC$. Since $C,E,F,X$ are concyclic, then $$\measuredangle CXF=\measuredangle AEF=\measuredangle ACB=\measuredangle CXA' \Longrightarrow X-F-A' \thickspace \text{collinear.}$$ By Pascal on $AACBXA'$ we get that $YZ\parallel BC$, which means $Y-Q-Z$ collinear. Since $\triangle AQZ$ and $\triangle ABC$ are homothetic, then they share their circumcircles are tangent at $A$, and so $$YQ\cdot YZ=YA^2=YP^2=YX\cdot YB \Longrightarrow BXQZ \thickspace \text{cyclic} \Longrightarrow FQ\cdot FB=FX\cdot FZ$$ Performing an inversion centered at $Y$, points $A$ and $P$ would remain fixed; it carries $Q$ to $Z$ and $X$ to $B$. Since $A-Q-P-B$ are collinear, then $AZPXY$ is cyclic, so $$FX\cdot FZ=FA\cdot FP$$ Therefore $$FQ\cdot FB=FX\cdot FZ=FA\cdot FP \Longrightarrow FQ=FP,$$ as desired.

Define some new points: 1. $Z = \overline{EF} \cap \overline{AY}$ 2. $P'$ is the other intersection of $\overline{CX}$ and the circumcircle of $\triangle APY$. Claim: $APYX$ are cyclic. Proof: Note that $\angle YPX = \angle YBP$ because of power of point ($PY^2 = AY^2 = YX \cdot YB$). We now do a simple angle chase: $\angle YAX = \angle ACX$ and $\angle YPX = \angle YBP = \angle XBA = \angle ACX$. So, $\angle YAX = \angle YPX$. Claim: $C,P',X,Z$ are collinear. Proof: Apply radax theorem on the circumcircles of $\triangle AFE,\triangle ABC,\triangle CEF$ Claim: $PP' \parallel BC$ Proof: $\angle XP'P = \angle XAP = \angle XAB = \angle XCB$ Claim: $YP' \parallel AC$ Proof: $\angle YP'Z = \angle YP'X = \angle YAX = \angle ACX = \angle ACZ$ We note that $\frac{ZP'}{ZC} = \frac{FP}{FB} = \frac{ZY}{ZA} = \frac{FQ}{FA} = \frac{FQ}{FB}$. Therefore, $FQ = FP$. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(25cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.77791256197981, xmax = 18.465449665946593, ymin = -4.0316249160484965, ymax = 13.42689090521705; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen ccqqqq = rgb(0.8,0,0); pen ffvvqq = rgb(1,0.3333333333333333,0); /* draw figures */ draw(circle((7,4.339899853470883), 8.236184234107235), linewidth(2) + blue); draw((2.578738377140784,11.288797348785437)--(14,0), linewidth(2) + blue); draw((2.578738377140784,11.288797348785437)--(0,0), linewidth(2) + blue); draw((xmin, 0*xmin + 0)--(xmax, 0*xmax + 0), linewidth(2) + blue); /* line */ draw(circle((4.789369188570392,-3.6076901985800918), 9.891973944233138), linewidth(2)); draw((xmin, -3.429874191921271*xmin + 0)--(xmax, -3.429874191921271*xmax + 0), linewidth(2) + qqwuqq); /* line */ draw((xmin, 0.6362536828094455*xmin + 9.64806555932756)--(xmax, 0.6362536828094455*xmax + 9.64806555932756), linewidth(2) + fuqqzz); /* line */ draw((xmin, 0*xmin + 8.13836949633405)--(xmax, 0*xmax + 8.13836949633405), linewidth(2) + ccqqqq); /* line */ draw((-2.372789507995125,8.13836949633405)--(xmax, -1.612940444789682*xmax + 4.311201331916102), linewidth(2) + fuqqzz); /* ray */ draw((xmin, -0.2781510869271784*xmin + 3.894115216980497)--(xmax, -0.2781510869271784*xmax + 3.894115216980497), linewidth(2) + ffvvqq); /* line */ draw((xmin, 0*xmin + 5.644398674392718)--(xmax, 0*xmax + 5.644398674392718), linewidth(2) + blue); /* line */ draw(circle((1.6966721339411817,7.208768531223929), 4.174287485251179), linewidth(2)); draw((xmin, 0*xmin + 3.1504278524513922)--(xmax, 0*xmax + 3.1504278524513922), linewidth(2) + ccqqqq); /* line */ draw((xmin, -0.9884019578179827*xmin + 5.793099701141701)--(xmax, -0.9884019578179827*xmax + 5.793099701141701), linewidth(2) + blue); /* line */ /* dots and labels */ dot((2.578738377140784,11.288797348785437),dotstyle); label("$A$", (2.659159316725142,11.48093646036516), NE * labelscalefactor); dot((0,0),dotstyle); label("$B$", (-0.2964884721536658,-0.36001275595058996), NE * labelscalefactor); dot((14,0),dotstyle); label("$C$", (14.188021499432542,-0.34165469515010044), NE * labelscalefactor); dot((8.289369188570392,5.644398674392718),linewidth(4pt) + dotstyle); label("$E$", (8.368516225677373,5.7899376122134045), NE * labelscalefactor); dot((1.289369188570392,5.644398674392718),linewidth(4pt) + dotstyle); label("$F$", (1.1354402702845143,5.808295673013894), NE * labelscalefactor); dot((-1.2355511849407201,4.2377851220259215),linewidth(4pt) + dotstyle); label("$X$", (-1.7100591537913565,4.3947249913762), NE * labelscalefactor); dot((-2.372789507995125,8.13836949633405),linewidth(4pt) + dotstyle); label("$Y$", (-2.7197524978182783,8.36006612428194), NE * labelscalefactor); dot((0.719662950491684,3.150427852451391),dotstyle); label("$P$", (0.786637115075214,3.3299574649478063), NE * labelscalefactor); dot((1.8590754266491012,8.13836949633405),linewidth(4pt) + dotstyle); label("$Q$", (1.9248368847055624,8.28663388107998), NE * labelscalefactor); dot((-6.292563788796045,5.644398674392718),linewidth(4pt) + dotstyle); label("$Z$", (-6.226142110711771,5.7899376122134045), NE * labelscalefactor); dot((2.67368131739068,3.150427852451389),linewidth(4pt) + dotstyle); label("$P'$", (2.7509496207275896,3.2932413433468275), NE * labelscalefactor); label("$P'Y$", (-7.401058001943098,12.986297446005302), NE * labelscalefactor,blue); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Let $S = XF \cap AC$, $R = XF \cap \Gamma \neq X$. Since $\measuredangle YPA = -\measuredangle YAP = -\measuredangle YAB = \measuredangle YXA$, so $(APXY)$. Also, by Reim's Theorem on $AEC$ and $RFX$, we get $AR \parallel BC$, so $$\measuredangle YAS = \measuredangle ABC = \measuredangle BCR = \measuredangle BXR,$$ so $(ASPXY)$, and by Reim's Theorem again on $ASC$ and $BXY$, we get $SY \parallel BC$, so $S \in YQ$. Finally, note that $\Delta XPB \stackrel{-}\sim \Delta ESF$ and $\Delta XPF \stackrel{-}\sim \Delta ASF$, so $$\frac{AQ}{QF} = \frac{AS}{SE} = \frac{AS}{SF} \cdot \frac{SF}{SE} = \frac{XP}{PF} \cdot \frac{PB}{PX} = \frac{BP}{PF}$$ where lengths are directed as necessary. As $F$ is the midpoint of $AB$, this finishes the problem.

Aprovecho para dejar otro problema: Desde un punto P exterior a una circunferencia, se trazan las tangentes PC y PD, Formando entre si un ángulo recto. Luego se traza la secante PAB, de modo que el arco BC es el doble del arco AC. Hallar la medida del ángulo BPC. Entonces trazamos BC, AC, AD, CD y sea O el centro de la cía. Entonces trazamos tb OC y OD. ⟹ ∠OCP=∠ODP=∠CPD=90°…(θ) ⟹Sabemos que (BC) ̂=2(AC) ̂ ⟹∠PBC=x y ∠BAC=2x Por ángulo semiinscripto ∠ACP=x ⟹Por ángulo externo en ∆ACP: ∠BPC=x ⟹∠CAP=180°−2x Luego:∠APD=90°−x ⟹Por pot. de un punto PC=PD. ⟹∆CDP triángulo rectángulo isósceles. ⟹∠CDP=∠DCP=45° Entonces por (θ) y por ser complementos: ∠ODC=∠OCD=45° Por ser suplemento ∠COD=90° ⟹ ∆CDO es triángulo rectangulo isósceles. ⟹Por arco capaz ∠CBD=45° y entonces ∠PBD=45°−x Por semiinscripto ∠ADP=45°−x Por se suplemento ∠DAP=45°+2x Aplicamos T. del Seno en ∆ACP Y ∆APD: AP/sin⁡x =PC/sin⁡〖180°−2x〗 ...(1) AP/sin⁡〖45°−x〗 =PD/sin⁡〖45°+2x〗 …(2) Sabemos que PC=PD ⟹Por lo anterior en (1) y (2) tenemos lo siguiente: sin⁡〖180°−2x〗/sin⁡x =sin⁡〖45°+2x〗/sin⁡〖45°−x〗 ⟹Aplicando transformaciones: sin⁡2x/sin⁡x =(〖√2/2(sin〗⁡2x+cos⁡2x))/(√2/2(cos⁡x−sin⁡x))⟹(2 sin⁡x cos⁡x)/sin⁡x =(sin⁡2x+cos⁡2x)/(cos⁡x−sin⁡x ) ⟹2〖cos⁡x〗^2−2 sin⁡x cos⁡x=sin⁡2x+2〖cos⁡x〗^2−1 ⟹−sin⁡2x=sin⁡2x−1⟹2sin⁡2x=1⟹sin⁡2x=1/2 Entonces 2x=150° o 2x=30°. Notemos que si pasa lo primero ∠PAD=195° Lo cual no puede pasar ∵La suma de los ángulos internos de un triángulo es 180° Entonces 2x=30° ⟹x=15° ∴∠BPC=15°∎