This is also Grade 11 P5.

Nice projective problem indeed

This also works :

Probably a trivial extension of this problem : "If $S$ is the projection of $E$ on $BC$ and $T$ is the midpoint of $FG$, then $ASOT$ is cyclic."

Let $E'$ be the antipode of $E$ w.r.t. $\odot O$. Clearly $A, G, E'$ are colinear so quadrilateral $EGFE'$ is harmonic. Taking pencil at $F$, rotate it $90^{\circ}$, translate it to $O$ and project it to $AC$ gives $$-1 = F(E, F; G, E') = O(A, F; K, C) = (A, F; K, C)$$Moreover, by taking pencil at $B$, $$-1 = B(A, F; K, C) = (E, F; BK\cap EF, {\infty}_{EF}) = -1$$hence $BK$ bisects $EF$ as desired.

Redefine $K = BM \cap AC$ where $M$ is the midpoint of $EF$. Let the tangent to $\odot(O)$ at $K$, other than $\overline{AC}$ meet $\odot(O)$ and $\overline{AB}$ at $X$ and $Y$, respectively. Note that if $P_{\infty}$ denotes the point at infinity on $\overline{AC}$ then $BP_{\infty}$ is tangent to $\odot(O)$. By Brianchon's theorem, $\overline{YM} \parallel \overline{AC}$ so $AE = EY = XY \implies AXE = 90^\circ$ so $X = G$ and $BK$ bisects $EF$ as required.

Denote $EM\cap \Gamma =E'$. It is clear that $E'$ lies on $AG$, as $\angle EGE'=90$. Also, define $EK\cap \Gamma=K'$, $EB\cap \Gamma=B'$. Note that, by symmedians, we have that quadrilaterals $GFE'E$, $FB'E'E$, and $FK'GE$ are all harmonic. By Ptolemy's, $K'E=\frac{FK'\cdot GE+K'G\cdot EF}{GF}=\frac{2EF\cdot K'G}{GF}$, as product of opposite sides are equal in harmonic quads. Similarly, $B'E=\frac{2B'E'\cdot EF}{E'F}$, so $$\frac{K'E}{B'E}=\frac{K'G\cdot E'F}{B'E'\cdot FG}=\frac{K'G}{B'E'}\cdot\frac{E'E}{GE}=\frac{K'F}{EF}\cdot\frac{EF}{FB'}=\frac{FK'}{FB'}$$Therefore, $FK'EB'$ is harmonic. Now, $$(FE;B'K')\stackrel{E}{=}(FA;BK)\stackrel{C}{=}(FE;P_\infty,KC\cap EF)$$as desired.

There is also a relatively short solution using complex numbers. WLOG, we may assume that $\odot O$ is the unit circle and $BC$ lies over the real axis. It is immediately obvious that $f=\tfrac{-1}{e}$ and $b=-c$. Then, $$a=\frac{-2e}{e^2-1}\quad \text{and}\quad b=-c=\frac{2e}{e^2+1}$$Since $\angle EGA=90^{\circ}$, $$\frac{g+\frac{2e}{e^2-1}}{\frac{1}{g}+\frac{2e}{e^2-1}}=ge\Rightarrow g=\frac{e(e^2-3)}{3e^2-1}$$and $$k=\frac{\frac{-e^2+3}{3e^2-1}}{\frac{-1}{e}+\frac{e(e^2-3)}{3e^2-1}}=\frac{e(-e^2+3)}{e^4-6e^2+1}$$By computation, $BK$ bisects $EF$ if and only if $$\frac{(e^4-6e^2+1)(e^4-4e^2-1)}{(e^2+1)(e^6-7e^4+7e^2-1)}=\overline{\left(\frac{(e^4-6e^2+1)(e^4-4e^2-1)}{(e^2+1)(e^6-7e^4+7e^2-1)}\right)}$$which is true. $\blacksquare$

Extend $AG$ to meet $\Gamma$ at $E'$. Then, $\angle E'GE = 90^{\circ}$, so $E', M,$ and $E$ collinear. Now since $AB = AC$, $E'$ is the reflection of $F$ across $BC$, so $BE'$ is tangent to $\Gamma$. Let $E'K$ meet $\Gamma$ at $L$. We also know that $E'FLG$ is harmonic, so $-1 = (GF;LE') \stackrel E' = (AF;KB)$. Let $CK$ meet $EF$ are $M$. We know that $-1 = (AF;KB) \stackrel C = (EF;M \infty)$, which implies $M$ is the midpoint of $EF$.

[asy][asy] import markers; unitsize(2inch); pair A, B, C, O, E, F, G, J, K, L, X, Y, M, N; A = dir(90); B = dir(205); C = dir(335); draw(B--A--C); draw(B--C, lightblue); O = .5B + .5C; E = foot(O, A, B); F = foot(O, A, C); draw(E--F, lightblue); X = .5E + .5F; J = 2*foot(F, B, C) - F; L = 2*foot(E, B, C) - E; G = extension(A, J, X, L); draw(circumcircle(A, G, E), dashed+lightmagenta); draw(circumcircle(E, F, J), dashed+lightmagenta); draw(A--O); draw(E--G); Y = extension(E, G, A, X); K = extension(.5A+.5E, G, A, F); draw(K--B, dotted); draw(G--L, lightred); draw(G--F); draw(L--O); M = .5A+.5E; draw(G--M--X, green); draw(G--K); draw(K--Y, lightblue); draw(K--O, lightred); N = extension(K, O, F, X); draw(G--O--F, green); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$E$", E, dir(200)); dot("$F$", F, dir(F)); dot("$O$", O, dir(O)); dot("$X$", X, dir(315)); dot("$G$", G, dir(40)); dot("$Y$", Y, dir(135)); dot("$K$", K, dir(K)); dot("$M$", M, dir(M)); dot("$N$", N, dir(310)); [/asy][/asy] Let the midpoint of $EF$ be $X$ and let $EG\cap AX = Y$. Since $OE$ and $OG$ are tangents to $(AGE)$, $(A, X; O, Y) = -1$ and $M, G, K$ are collinear. From $\angle GOF = 2\angle GEF = \angle GMX$, $\angle OGF = \angle MGX\implies \angle XGF = \angle MGO = 90^{\circ}$. Then $GX$ and $FO$ concur on $\odot O$, and a homothety at $F$ by factor $1/2$ shows that $KO$ bisects $XF$. Call $N$ the midpoint of $XF$. Since $-1 = (A, X; O, Y) \stackrel{K}{=} (F, X; N, KY\cap EF)$, $KY$ is parallel to $EF$. Finally, $-1 = (A, X; O, Y) \stackrel{K}{=} (C, KX\cap BC; M, P_{\infty})$, so $KX\cap BC = B$.

Let $N=EF\cap CK$; we want to prove $(E, F; N, P_{\infty})=-1$. But we have $(E, F; N, P_{\infty})\overset{C}=(A, F; K, B)\overset{G}=(L, F; G, X)$, where $L=AG\cap \omega, X=BG\cap \omega$. And note $L, E$ are reflections over $M$, so thus $BL$ tangent to $\omega$, meaning $GFXL$ is harmonic and we're done.

Let $M$ be the midpoint of $\overline{BC}$, $N$ be the midpoint of $\overline{AE}$, $E'$ be the antipode of $E$, $X = \overline{BE'} \cap \overline{FE}$ and $Y = \overline{KC} \cap \overline{FE}$. Note that $X$ is on the polar of $A$ and thus $\overline{E'A}$ is the polar of $X$. $\angle AGE + \angle EGE' = 180$ so $A,G,E'$ are collinear and the tangent at $G$ passes through $X$. The tangent at $G$ also passes through $N$ so $X,K,G,N$ are collinear. $(X, Y; F, E) \overset{K}= (N,C;A,E) \implies \frac{XF}{XE} \frac{YE}{YF} = \frac{CE}{CA} \implies \frac{YE}{YF} = \frac{CE}{XF} \frac{XE}{CA}$. Notice that $CE = BF$ and $XE=BC$. So, we have $\frac{YE}{YF} = \frac{BF}{XF} \cdot \frac{BC}{CA} = -1$ by similar triangles. Thus, $YE = -YF$ and we are done.

Let the parallel line at $A$ to $BC$ intersect the perpendicular line from $E$ to $BC$ at $X, D=AO\cap BK$ Claim: $X,G,F$ are collinear Proof. Note that $A,X,E,G$ are cocyclic So $\angle{AGX}=\angle{AEX}=90-\angle{AEF}=\frac12\hat{A}$ $\angle{AGF}=360-90-\angle{EGF}=90+\hat{B}=180-\frac12\hat{A} \quad \blacksquare$ $Polar(B)=EX$ $Polar(K)=GF$ hence $Polar(BK)=X$ $Polar(AO)=P_\infty$ So $Polar(D)=XP_\infty$ Wich passes from $A$ Hence $D\in Polar(A)=EF$ and immediately $D$ is the midpoint of $EF$ as desired.

Let $AG$ intersect $\Gamma$ at $F'$, and let $M'$ be $CK \cap EF$. Then \[\angle F'GE = \angle AGE = 90^{\circ},\] so $EF'$ is a diameter, which implies that $\angle EFF' = 90^{\circ}$. Then since $EF || BC$, this implies that $F'$ is the reflection of $F$ across $BC$, which implies that $BF'$ is tangent. Now, if we let $T$ be $F'K \cap \Gamma$, we know that $F'FTG$ is a harmonic quadrilateral since the intersection of the tangents at $F$ and $G$ is $K$. Then \[-1 = (G, F; T, F') \stackrel{F'}{=} (A, F; K, B) \stackrel{C}{=} (E, F; M', \infty) \] so $M'$ is the midpoint of $EF$, as desired. $\square$

Let $E'$ be the antipode of $E$ in $\Gamma$. Let $L \neq E'$ be an intersection of $\overline{E'K}$ with $\Gamma$. By a well-known lemma, $E'FLG$ is harmonic. Since $\angle AGE=\angle EGE'=90^\circ$, we find that $A$, $G$, and $E'$ are collinear. Also, we have $\triangle BE'M \cong \triangle CEM$, so $\overline{BE'}$ is tangent to $\Gamma$. Let $\overline{EF}$ and $\overline{CK}$ intersect at $N$, and let $\infty$ be the point at infinity on $\overline{BC}$. Then, we have $$-1=(FG;E'L)\overset{E'}{=}(FA;BK)\overset{C}{=}(FE,\infty N),$$so $N$ is the midpoint of $\overline{EF}$.

Solved with L567, minakshee, Psyduck909, Pranav1056, KilleR_1234. Consider $EF \cap KC = D$ $\textcolor{red}{Claim:}$$(A,F;K,B)=-1$ $\textcolor{blue}{Proof:}$ Let $AG \cap \Gamma = Z$ and $KZ \cap \Gamma = X$. So now we know $XFZG$ is harmonic. Taking perspectivity at $Z$ gives, \[(G,F;X,Z) \stackrel{Z}{=} (A,F;K,B) = -1\] Take $FE \cap BC = P_{\infty}$ . Now to finish we take perspectivity at $C$ giving us : \[(A,F;K,B) \stackrel{C}{=} (E,F;D,P_{\infty})=-1\]which implies that $D$ is the midpoint of $EF$, as desired.

Oof, I had thought about $Z$ and $X$ motivated my so many tangents but did not think about taking perspectivity at $Z$ Once you discover that $EF$ and $BC$ are parallel, Projective is quite motivated (EGMO )

Let $E'$ be the antipode of $E$ with respect to $\omega$ and note that $E'$ lies on $\overline{AG}$ as $\angle E'EA=\angle AGE=90.$ Let $f(P,\ell)$ be the reflection of $P$ over $\ell.$ Since $E'=f(f(E,\overline{AO}),\overline{BC})$ and $F=f(E,\overline{AO}),$ we know $E'=f(F,\overline{BC}).$ Hence, $\overline{CE'}$ is tangent to $\omega$ at $E'.$ Therefore, $$-1=(G,F;\overline{FG}\cap\overline{KE'},E')\stackrel{E'}=(A,F;K,C)\stackrel{B}=(E,F;M,P_{\infty})$$where $P_{\infty}$ is the point at infinity along $\overline{BC}.$ Noticing $\overline{BC}\parallel\overline{EF}$ finishes. $\square$

[asy][asy] size(7cm); defaultpen(0.8); defaultpen(fontsize(9pt)); dotfactor*=0.6; pen greenfill,greendraw,turquoisedraw,lightbluedraw,bluedraw,purpledraw,pinkdraw; greenfill = RGB(204,255,204); greendraw = RGB(0,187,0); turquoisedraw = RGB(0,170,153); lightbluedraw = RGB(68,204,255); bluedraw = RGB(0,102,255); purpledraw = RGB(170,34,255); pinkdraw = RGB(255,17,255); pair A,B,C,M,E,F,EE,G,K,X,L; A = (0,3); B = (-2,0); C = (2,0); M = (B+C)/2; E = foot(M,A,B); F = foot(M,A,C); path c = circle(M,distance(M,E)); EE = 2M-E; G = intersectionpoints(c,A--EE)[0]; K = extension(M,foot(M,F,G),A,C); X = extension(B,K,E,F); L = intersectionpoints(c,EE--K)[0]; filldraw(c,greenfill,greendraw); draw(A--B--C--cycle,bluedraw); draw(A--EE,pinkdraw); draw(E--EE,turquoisedraw); draw(E--F,purpledraw); draw(B--K,lightbluedraw); draw(G--K,lightbluedraw); draw(EE--K,lightbluedraw); draw(C--EE,bluedraw); dot("$A$",A,dir(90)); dot("$B$",B,dir(180)); dot("$C$",C,dir(0)); dot("$M$",M,dir(240)); dot("$E$",E,dir(140)); dot("$F$",F,dir(40)); dot("$E'$",EE,dir(320)); dot("$G$",G,dir(50)); dot("$K$",K,dir(50)); dot("$X$",X,dir(110)); dot("$L$",L,dir(215)); [/asy][/asy] Let $M$ be the midpoint of $\overline{BC}$, let $\overline{BK}$ and $\overline{EF}$ intersect at $X$, and let $E'$ be the reflection of $E$ over $M$. We have $\angle AGE=\angle EGE'=90^\circ$, so $A$, $G$, and $E'$ are collinear. Since $\triangle BEM \cong \triangle CE'M$, $\overline{CE'}$ is tangent to $\omega$. Let $\overline{KE'}$ intersect $\omega$ at $L \ne E'$ and let $\infty_{BC}$ be the point at infinity on $\overline{BC}$. We have \[-1=(G,F;L,E')\overset{E'}{=}(A,F;K,C)\overset{B}{=}(E,F;X,\infty_{BC}),\]so $X$ is the midpoint of $\overline{EF}$. $\square$

Nice problem. Let $N$ be the midpoint of $AE$, $D=AG\cap\Gamma$ and $T=GG\cap EF$. Now $\angle AGE=90^\circ\implies N$ is the center of $\odot(AGE)\implies NE=NG$ and as $NE$ is tangent to $\Gamma$, we get that $NG$ is also tangent to $\Gamma$. Thus this gives that $\overline{K-G-N}$ are collinear. Now as $(G,D;E,F)=-1$, this means that $TD$ is tangent to $\Gamma$. Also, $AG\cdot AD=AE^2\implies\angle AED=\angle AGE=90^\circ\implies\overline{E-M-D}$ are collinear. Finally, $TD\perp DM\implies TD\parallel AC$. Now as $M$ is the midpoint of both $BC$ and $ED$, we get that $BDCE$ is a parallelogram, which further gives that $BD\parallel AC\implies\overline{T-B-D}$ are collinear. Thus we finally have, \begin{align*} -1&=(A,E;N,\infty_{AC})\\ &\overset{T}{=}(A,F;K,B)\\ &\overset{C}{=}(E,F;BC\cap EF,\infty_{BC}) ,\end{align*} which thus implies that $BC\cap EF$ is the midpoint of $EF$ and we are done.

Had a version that replaced $\odot O$ with $\Gamma$. Let $P_{\infty}$ denote the point at infinity along lines $EF$ and $BC$, and $X$ denote the $E$-antipode wrt $\Gamma$. Then since $\angle EGX = 90^{\circ}$, points $A$, $G$, and $X$ are collinear. Moreover, since $\Gamma$ is centered at the midpoint of $\overline{BC}$, and $E$ and $F$ are reflections across the perpendicular bisector of $\overline{BC}$, points $X$ and $F$ are reflections across $BC$. So, $\overline{FX}$ is tangent to $\Gamma$. Then to finish, note that $$-1 = (F, G; \overline{KX} \cap \Gamma, X) \overset{X}{=} (F, A; K, B) \overset{C}{=} (F, E; \overline{CK} \cap \overline{EF}, P_{\infty}),$$which implies that $\overline{CK}$ bisects $\overline{EF}$, as needed.

Rename $\odot O$ to $\Gamma$. Also swap $E$ and $F$ in terms of definition. We present two distinct solutions, one with multiplicity 2. Solution 1 (synthetic): We will need the following claim, which will be proven in two different ways. Claim: Let $ABCD$ be a convex cyclic quadrilateral. Then $\overline{AA} \cap \overline{BB}$, $\overline{CC} \cap \overline{DD}$, and $\overline{AC} \cap \overline{BD}$ are collinear. Proof 1: This is purely projective, so employ a projective transformation sending $\overline{AC} \cap \overline{BD}$ to the center of $(ABCD)$, hence $ABCD$ is now a rectangle and the conclusion is obvious. $\blacksquare$ Proof 2: Let $W=\overline{AA} \cap \overline{DD}$, and define $X,Y,Z$ in cyclic fashion. By Brianchon on $WAXYCZ$, we get that $\overline{WY}, \overline{XZ}, \overline{AC}$ are concurrent. By Brianchon on $WXBYZD$, we get that $\overline{WY}, \overline{XZ}, \overline{BD}$ are concurrent. Hence $\overline{AC} \cap \overline{BD}$ lies on $\overline{WY}$ as desired. $\blacksquare$ For the main problem, let $E'$ and $F'$ be the antipodes of $E$ and $F$ respectively, so $G$ lies on $\overline{AE'}$, let $N$ be the midpoint of $\overline{EF}$, and let $\overline{EN}$ intersect $\Gamma$ again at $H$. Since $EGFE'$ is harmonic, we have $$-1=(E,F;E'G)\stackrel{H}{=}(E,F;N,\overline{HG} \cap \overline{EF}),$$hence $\overline{GH} \parallel \overline{EF}$. Therefore, due to symmetry, $\overline{F'G}$ passes through $N$ as well. Now apply out claim to $FGEF'$ to get that $C$, $K$, and $N$ are collinear, as desired. $\blacksquare$ Solution 2 (coordinates): First we change the setup of the problem slightly. Define $E'$ as before, so $G$ lies on $\overline{AE'}$, and once again let $N$ be the midpoint of $\overline{EF}$. Let $K'=\overline{CN} \cap \overline{FF}=\overline{CN} \cap \overline{AB}$. We are going to show that the point $G' \neq F$ such that $\overline{K'G'}$ is tangent to $\Gamma$ is $G$ by showing it lies on $\overline{AE}$, which clearly implies $K'=K$ and hence solves the problem. Place the problem in the coordinate plane with $A=(0,a),B=(-1,0),C=(1,0)$, so the midpoint $M$ of $\overline{BC}$ is the origin. Lines $\overline{AB}$ and $\overline{AC}$ have equations $y=ax+a$ and $y=-ax+a$ respectively, so we can easily calculate $F=(-\tfrac{a^2}{a^2+1},\tfrac{a}{a^2+1})$. It is clear that $E'$ is the reflection of $F$ over $\overline{BC}$, hence $E'=(\tfrac{-a^2}{a^2+1},-\tfrac{a}{a^2+1})$. Furthermore, $N=(0,\tfrac{a}{a^2+1})$. To calculate $G'$, we do not actually have to calculate $K'$ (although this is not hard either): instead, notice that $\overline{K'M}$ should bisect $\overline{FN}$ for homothety reasons, hence its slope is twice that of $\overline{MF}$, so its equation is $y=-\tfrac{2}{a}x$. Now, $G'$ is just the reflection of $F$ over $\overline{MK}$, whose equation we know. The foot from $F$ to $\overline{MK}$ can be found by intersecting $\overline{MK}$ with the line that has slope $\tfrac{a}{2}$ passing through $F$, and can be calculated as $(-\tfrac{a^4+2a^2}{(a^2+1)(a^2+4)},\tfrac{2a^3+4a}{(a^2+1)(a^2+4)})$. Since this point is the midpoint of $\overline{FG'}$, it folows that $G'=(-\tfrac{a^4-4a^2+4a}{(a^2+1)(a^2+4)},\tfrac{3a^3+4a}{(a^2+1)(a^2+4)})$. To show that $A,G',E'$ are collinear, we calculate the slope of $\overline{AE'}$, which is just $$\frac{a+\frac{a}{a^2+1}}{\frac{a^2}{a^2+1}}=\frac{a^2+2}{a}.$$We will now calculate the slope of $\overline{G'E'}$, since by scaling all the coordinates by $\tfrac{a^2+1}{a}$ beforehand this is just equal to $$\frac{\frac{3a^2+4}{a^2+4}+1}{-\frac{a^3}{a^2+4}+a}=\frac{\frac{4a^2+8}{a^2+4}}{\frac{4a}{a^2+4}}=\frac{a^2+2}{a},$$hence the desired collinearity holds and we are done. $\blacksquare$

Let $E'$ be the intersection of $\overline{AG}$ and $\Gamma$. Claim: $B$ is tangent to $\Gamma$ at $E'$. Proof. Note that $EE'$ is a diameter of $\Gamma$. Angle chase to get \[ \measuredangle FE'M = \measuredangle FE'E = \measuredangle FEE' + \measuredangle E'FE = \measuredangle BFE' + \measuredangle E'FE = \measuredangle BFE = \measuredangle FBM. \]$\blacksquare$ As such, \[ (G,F;\overline{KE'} \cap \Gamma,E') \overset{E'}= (AF;KB) \overset{C}= (EF;I\infty) = -1 \]and the result follows.

Let $X\neq G$ be the intersection of $AG$ and $\Gamma$. Since $A$, $G$, and $X$ are collinear in that order, we find that $\angle XGE=180^\circ-\angle AGE=90^\circ$, implying that $XE$ is a diameter of $\Gamma$. It thus follows that $\triangle BXM\cong\triangle CEM$, so that $\angle BXM=\angle CEM=90^\circ$ as $CE$ is tangent to $\Gamma$ at $E$. Thus $BX$ is tangent to $\Gamma$ at $X$. Now let $L\neq X$ be the intersection of $KX$ and $\Gamma$. Since $FK$ and $GK$ are tangents to $\Gamma$ and $X$ lies on $KL$, it follows that $FXGK$ is harmonic, so that $(GF;LX)=-1$. Let $N=EF\cap CK$. Projecting, we then find that \begin{align*} (EF;N\infty)&\overset{C}=(AF;KB)\\ &\overset{X}=(GF;LX)\\ &=-1. \end{align*}This forces $N$ to be the midpoint of $EF$, implying that $CK$ indeed bisects $EF$. $\blacksquare$ - Jörg

This problem was copied from APMO 2016 P3 https://artofproblemsolving.com/community/c6h1243425p6362864

Denote the midpoint of $EF$ as $N$ and $H=AG \cap \Gamma$. We make a few synthetic observations: $H$ is simply the antipode of $E$ on $\Gamma$. $\triangle EMN$ and $\triangle EHF$ are homothetic about $E$, so $H$ is also the reflection of $F$ over $BC$. Since $\triangle KFN$ and $\triangle KBC$ are homothetic about $K$, it remains to show $KM$ bisects $FN$. Thus we finish by noting the harmonic bundle \[-1 = (H, HK \cap \Gamma; FG) \overset{H}{=} (BK, FA) \overset{M}{=} (\infty, KM \cap FN; FN). \quad \blacksquare\]

Let $\overline{AG} \cap \Gamma \neq G = E'$. Then notice that $E'$ is the antipode of $E$. Also, let $\overline{KE'} \cap \Gamma \neq E' = P$. Then $FPGE'$ is a harmonic quadrilateral, so $-1 = (F, G; P, E') \overset{E'}= (F, A; K, B)$. Notice that since $AB = AC$, $EF \parallel BC$ so $-1 = (F, A; K, B) \overset{C}= (F, E; \overline{CK} \cap \overline{EF}, P_{\infty})$ so $CK$ bisects $EF$ as desired.

Let $KC \cap FE = X$. Let $AG \cap \Gamma = E'$. Let $KE' \cap \Gamma = W$. Now from KF and KG tangents, we get that $(F,G;W,E') = -1$. Now projecting trough E', we have that $(F,G;W,E')\stackrel{E'}{=}(F,A;K,B) = -1$. Now projecting trough C, we have that $(F,A;K,B)\stackrel{C}{=}(F,E;X,P_\infty) = -1$. From AB = AC, we get that $FE \parallel BC$ $\Rightarrow$ if L is the midpoint of FE, then $(F,E;L,P_\infty) = -1$ and since $(F,E;X,P_\infty) = -1$ its obvious that $L \equiv X$ $\Rightarrow$ X is the midpoint of FE $\Rightarrow$ CK bisects EF, which is what we wanted to prove. We are ready.

Let $E'=AG \cap \Gamma$ and $E'K \cap \Gamma = P.$ Since $(F, G; P, E')=-1$, taking perspectivity at $E'$ and noting $BE'$ is tangent to $\Gamma$ gives $$(F, A; K, B)=-1.$$Finally taking perspectivity at $C$ gives $(F, E; CK \cap EF, P_{\infty})=-1$ which implies $CK \cap EF$ is the midpoint of $EF.$

Let $E' \neq G = \overline{AG} \cap \Gamma$, $D \neq E' = \overline{KE'} \cap \Gamma$, and $N = \overline{CK} \cap \overline{EF}$. Then, we have \[-1 = (F,G;D,E') \overset{E'}{=} (F,A;K,B) \overset{C}{=} (F,E;N, \infty). \ \square\]