Let $H_B$ be on $\Gamma$ such that $BH_B\perp AC$, let $G$ be the Miquel Point of $BCEF$, and let $K$ be the second intersection of $GE$ and $\Gamma$. Then \[\measuredangle GKB = \measuredangle GCB = \measuredangle GEF,\]as $\triangle GEF\sim\triangle GCB$ by spiral similarity. Hence $EF \parallel BK$. Now, noting that $GBPC$ is harmonic and letting $N = BQ\cap AC$, we get \[ -1 = (GP;BC) \overset{E}{=} (KQ;H_BA) \overset{B}= (\infty_{EF}N;EF), \]implying that $N$ is the midpoint of $\overline{EF}$ as desired. $\square$

tastymath75025 wrote: Let $ABC$ be an acute triangle with orthocenter $H$ and circumcircle $\Gamma$. Let $BH$ intersect $AC$ at $E$, and let $CH$ intersect $AB$ at $F$. Let $AH$ intersect $\Gamma$ again at $P \neq A$. Let $PE$ intersect $\Gamma$ again at $Q \neq P$. Prove that $BQ$ bisects segment $\overline{EF}$. Proposed by Luke Robitaille Let $M$ be the midpoint of $EF$ and $D$ be feet from $A$ on $BC$. We have, $\triangle HDE \sim FBE$ so \[\frac{HP}{HE}=\frac{2HD}{HE} = \frac{2FB}{FE} = \frac{FB}{FM}.\]This gives $\triangle FBM \sim \triangle HPE$, so $\angle ABM=\angle FBM = \angle HPE = \angle HPQ = \angle ABQ$ and we are done. $~\square$

Let $K$ be the point so that $(AK;BC)=-1$. It is well known that $KP$ and $EF$ intersect at some point $R$ on $BC$. Now, apply Pascal's theorem on the cyclic hexagon $(KPQBCA)$. We see $KP\cap BC = R$, $PQ\cap AC=E$, so $BQ\cap AK$ lies on $EF$. However, as $EF$ and $BC$ are anti-parallel in $\angle BAC$, the $A$-symmedian in $\Delta ABC$ is the $A$-median of $\Delta AEF$, and as such $AK\cap EF$ is the midpoint of $EF$, which $BQ$ thus passes through.

pieater314159 wrote: Let $K$ be the point so that $(AK;BC)=-1$. It is well known that $KP$ and $EF$ intersect at some point $R$ on $BC$. Now, apply Pascal's theorem on the cyclic hexagon $(KPQBCA)$. We see $KP\cap BC = R$, $PQ\cap AC=E$, so $BQ\cap AK$ lies on $EF$. However, as $EF$ and $BC$ are anti-parallel in $\angle BAC$, the $A$-symmedian in $\Delta ABC$ is the $A$-median of $\Delta AEF$, and as such $AK\cap EF$ is the midpoint of $EF$, which $BQ$ thus passes through. wait yes this is what I did

Let $R$ be the midpoint of $AH$. As $HR \cdot HP = HB \cdot HE = \frac 12 \text{Pow}(H)$ we have $B, R, E, P$ cyclic. Now since $\angle ABQ = \angle RPE = \angle RBE$ we have $R, Q$ isogonal wrt $\angle ABE$. But $AFHE$ is cyclic, and so since $BR$ is a median of $\triangle BAH$ we have $BQ$ is a median of similar $\triangle BEF$, as desired.

Why can't we have real geometry like this in the contest? tastymath75025 wrote: Let $ABC$ be an acute triangle with orthocenter $H$ and circumcircle $\Gamma$. Let $BH$ intersect $AC$ at $E$, and let $CH$ intersect $AB$ at $F$. Let $AH$ intersect $\Gamma$ again at $P \neq A$. Let $PE$ intersect $\Gamma$ again at $Q \neq P$. Prove that $BQ$ bisects segment $\overline{EF}$. Proposed by Luke Robitaille Let $T=(AEF) \cap \Gamma.$ Let $TE \cap \Gamma=D$ and $BE \cap \Gamma=Y.$ Then $\measuredangle TEF=\measuredangle TAB=\measuredangle TDB$ and so $EF \parallel BD.$ So, $$-1=(B,C;P,T) \overset{E}{=} (Y,A;Q,D) \overset{B}{=} (E,F; BQ \cap EF, \infty)$$so done. $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -20.892503579514237, xmax = 44.58250800041276, ymin = -23.82258916154907, ymax = 17.60163912792321; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); draw((-9.207161569498936,12.160844964529835)--(-13.65757092076063,-9.43064345595649)--(16.524055689306913,-8.775538382249595)--cycle, linewidth(0.4) + rvwvcq); /* draw figures */ draw((-9.207161569498936,12.160844964529835)--(-13.65757092076063,-9.43064345595649), linewidth(0.4) + rvwvcq); draw((-13.65757092076063,-9.43064345595649)--(16.524055689306913,-8.775538382249595), linewidth(0.4) + rvwvcq); draw((16.524055689306913,-8.775538382249595)--(-9.207161569498936,12.160844964529835), linewidth(0.4) + rvwvcq); draw(circle((1.262822751839856,-1.2516149962838792), 17.01513014612322), linewidth(0.4) + dtsfsf); draw(circle((-9.03674193706565,4.3093690417106725), 7.8533252204227315), linewidth(0.4) + linetype("2 2") + wrwrwr); draw((-9.207161569498936,12.160844964529835)--(-8.615328719291325,-15.105739920034928), linewidth(0.4) + wrwrwr); draw((-15.69490840146251,0.1448118803658121)--(12.818858446805242,11.2372877940415), linewidth(0.4) + wrwrwr); draw((6.237026976644165,15.02019947596998)--(-8.615328719291325,-15.105739920034928), linewidth(0.4) + wrwrwr); draw((-1.3146476639940998,5.739046297430744)--(-12.298037341938633,-2.8347662663502065), linewidth(0.4) + wrwrwr); draw((12.818858446805242,11.2372877940415)--(-13.65757092076063,-9.43064345595649), linewidth(0.4) + linetype("2 2") + wvvxds); draw((-1.3146476639940998,5.739046297430744)--(-13.65757092076063,-9.43064345595649), linewidth(0.4) + wrwrwr); draw((-12.298037341938633,-2.8347662663502065)--(16.524055689306913,-8.775538382249595), linewidth(0.4) + wrwrwr); draw((2.1875170149572605,15.738370124641593)--(-8.615328719291325,-15.105739920034928), linewidth(0.4) + wrwrwr); draw((6.237026976644165,15.02019947596998)--(-1.3146476639940998,5.739046297430744), linewidth(0.4) + wrwrwr); draw((-13.65757092076063,-9.43064345595649)--(2.1875170149572605,15.738370124641593), linewidth(0.4) + linetype("2 2") + wrwrwr); draw((-15.69490840146251,0.1448118803658121)--(-9.207161569498936,12.160844964529835), linewidth(0.4) + wrwrwr); /* dots and labels */ dot((-9.207161569498936,12.160844964529835),linewidth(4pt) + dotstyle); label("$A$", (-10.381878491140686,12.655462615747414), NE * labelscalefactor); dot((-13.65757092076063,-9.43064345595649),linewidth(4pt) + dotstyle); label("$B$", (-15.080746177707685,-10.900703023489811), NE * labelscalefactor); dot((16.524055689306913,-8.775538382249595),linewidth(4pt) + dotstyle); label("$C$", (17.007573945032746,-10.035122133859046), NE * labelscalefactor); dot((-8.866322304632368,-3.5421068811084915),linewidth(4pt) + dotstyle); label("$H$", (-10.691014523151672,-4.408846351259079), NE * labelscalefactor); dot((-1.3146476639940998,5.739046297430744),linewidth(4pt) + dotstyle); label("$E$", (-2.1588600396484363,6.967359626745249), NE * labelscalefactor); dot((-12.298037341938633,-2.8347662663502065),linewidth(4pt) + dotstyle); label("$F$", (-13.596893224054948,-3.8524014936393027), NE * labelscalefactor); dot((-8.615328719291325,-15.105739920034928),linewidth(4pt) + dotstyle); label("$P$", (-9.207161569498936,-16.897942044502962), NE * labelscalefactor); dot((-15.69490840146251,0.1448118803658121),linewidth(4pt) + dotstyle); label("$T$", (-17.059216782578,-0.5137323479206409), NE * labelscalefactor); dot((12.818858446805242,11.2372877940415),linewidth(4pt) + dotstyle); label("$D$", (13.050632735292115,11.728054519714451), NE * labelscalefactor); dot((6.237026976644165,15.02019947596998),linewidth(4pt) + dotstyle); label("$Y$", (6.496948856659194,15.499514110248496), NE * labelscalefactor); dot((2.1875170149572605,15.738370124641593),linewidth(4pt) + dotstyle); label("$Q$", (2.4163532341141685,16.241440587074866), NE * labelscalefactor); dot((-6.806342502966366,1.4521400155402702),linewidth(4pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Let $a,b,c\in \mathbb C$ be vertices of triangle $ABC$ and let the circumscribed circle be the unit circle centered at $0$: $|a|=|b|=|c|=1$. We have also $|p|=1=|q|$. Foot of $B$ on segment $AC$: $$2e=a+b+c-\frac{ac}{b}$$Foot of $C$ on segment $AB$: $$2f=a+b+c-\frac{ab}{c}$$$$P\in AH\implies AP\perp BC\iff$$$$\frac{p-a}{b-c}=-\frac{\frac{1}{p}-\frac{1}{a}}{\frac{1}{b}-\frac{1}{c}}$$$$p=-\frac{bc}{a}$$Points $P,Q,E$ are collinear: $$\frac{q-p}{e-p}=\frac{\frac{1}{q}-\frac{1}{p}}{\overline e-\frac{1}{p}}$$$$q=\frac{p-e}{p\overline e-1}=\frac{a^2c-abc-ab^2-a^2b-2b^2c}{b^2-ab-bc-ca-2a^2}\cdot \frac{a}{b}$$Two final medium hard computations $$\frac{e+f}{2}-b=(c-b)\cdot\frac{2bc-ac+ab}{4bc}$$$$q-b=\frac{(c-b)(a+b)(a-b)^2}{b(b^2-ab-bc-ca-2a^2)}=\frac{(c-b)(a-b)^2}{b(b-2a-c)}$$So $$\frac{q-b}{\frac{e+f}{2}-b}=\frac{4c(a-b)^2}{(2bc-ac+ab)(b-2a-c)}$$Now you see that $$\frac{q-b}{\frac{e+f}{2}-b}=\overline{\left(\frac{q-b}{\frac{e+f}{2}-b}\right)}$$Which means points $\frac{e+f}{2},b,q$ are collinear.

Let $S \in \odot O$ such that $SP \parallel BE$, let $H_B = BE \cap \odot O$. Then \[ -1 = (H,H_B; E,\infty ) \stackrel{P}{=} (A,H_B; Q,S)\]Note that by basic angle chasing $BS \parallel EF$, (for example by proving that they are both orthogonal to AO), thus \[-1= (A,H_B; Q,S) \stackrel{B}{=} (F,E; BQ\cap FE, \infty) \]Therefore $BQ\cap FE$ is the midpoint of $EF$.

Solution. Let $R=\overline{PF}\cap\Gamma,\ R\neq P$. By Pascal's theorem on $BACRPQ$ we get that $M=\overline{CR}\cap \overline{BQ}$ lies on $EF$. Let $L$ be the point where the ray $AM$ meets $\Gamma$; it suffices to show that $ABLC$ is a harmonic quadrilateral. Note that $$\angle MLB=\angle ALB=\angle ACB=\angle EFA=\angle MFA$$i.e. $BFML$ is cyclic. Similarly, $CEML$ is cyclic. Thus $$\angle FLE=\angle FBM+\angle MCE=\angle ABQ+\angle ACR=\angle RPQ=\angle FPE$$hence, $EFPL$ is a cyclic quadrilateral and according to 2017 Sharygin Geometry Olympiad Correspondence Round, P12, we conclude that $ABLC$ is harmonic, as desired. $\blacksquare$

Here's my solution: Let $AK$ be the $A$-symmedian, where $K \in \Gamma$. Also let $X$ be the $A$-Ex point, and $D$ be the foot of the $A$-altitude in $\triangle ABC$. Then, $$-1=(B,C;D,X_A) \overset{P}{=} (B,C;A,PX_A \cap \Gamma) \Rightarrow PX_A \cap \Gamma=K$$This means that $KP$ and $BC$ meet on line $EF$. Applying Pascal to $BQPKAC$, we get that $BQ$ and $AK$ meet on line $EF$. But, $AK$ passes through the midpoint of $EF$, giving that $BQ$ bisects $EF$, as desired. $\blacksquare$

Let $BH$ intersects $\odot(ABC)$ again at $H_B$ and $M$ be the midpoint of $EF$. Then $\triangle PHH_B\cup E\sim\triangle BFE\cup M$ so $\angle ABM = \angle HPE = \angle APQ = \angle ABQ$ so $B,Q,M$ are colinear.

This problem was posted before: https://artofproblemsolving.com/community/c6h386417

Here is a different proof. Let $M$ be midpoint of $EF$. Let $BM$ intersect $\Gamma$ again in $Q'$. We have to prove that $Q$ =$Q'$.For this it is sufficient to prove that $\angle BQ'E=90^{\circ}-B$ Let $X\in BC$ s.t. $EX\perp BC$. Claim: $MXCQ'$ is cyclic. Proof: Let $AH$ meet $BC$ in $D$. Let $D'$ be reflection of $D$ in $X$. $\triangle DED'$ is isoceles by construction. So $\angle ED'D=\angle EDD'=\angle EDC=A$. Also we have $\angle FDB=A$. Hence $FD\parallel ED'$. But $M$ and $X$ are midpoints of $EF$ and $DD'$ resp. So $MX \parallel FD$ which gives that $\angle MXB=A=\angle CQ'B$ and our claim is proven. So $BE^2=BX\cdot BC=BM\cdot BQ'$. Thus $\angle BQ'E=\angle BEM=\angle BEF=90^{\circ}-B$ completing the proof.

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This is off topic, but Luke writes problems?

mathapple101 wrote: This is off topic, but Luke writes problems? You don't?

Let the tangent at $A$ to $\Gamma$ be $\ell$ (note $\ell \parallel EF$), and let $BQ$ intersect $EF, AC, \ell$ at $M, L, K$ respectively. In addition, let $ED \cap AB = F'$ and the point at infinity along $EF$ be $P_{\infty}$. By Pascal on $PQBCAA$, we have that $E, K, D$ are collinear. Then $$-1 = (AB;FF') \overset{E}{=} (LB;MK) \overset{A}{=} (EF;MP_{\infty})$$as desired.

Wizard_32 wrote: Why can't we have real geometry like this in the contest? tastymath75025 wrote: Let $ABC$ be an acute triangle with orthocenter $H$ and circumcircle $\Gamma$. Let $BH$ intersect $AC$ at $E$, and let $CH$ intersect $AB$ at $F$. Let $AH$ intersect $\Gamma$ again at $P \neq A$. Let $PE$ intersect $\Gamma$ again at $Q \neq P$. Prove that $BQ$ bisects segment $\overline{EF}$. Proposed by Luke Robitaille Let $T=(AEF) \cap \Gamma.$ Let $TE \cap \Gamma=D$ and $BE \cap \Gamma=Y.$ Then $\measuredangle TEF=\measuredangle TAB=\measuredangle TDB$ and so $EF \parallel BD.$ So, $$-1=(B,C;P,T) \overset{E}{=} (Y,A;Q,D) \overset{B}{=} (E,F; BQ \cap EF, \infty)$$so done. $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -20.892503579514237, xmax = 44.58250800041276, ymin = -23.82258916154907, ymax = 17.60163912792321; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); draw((-9.207161569498936,12.160844964529835)--(-13.65757092076063,-9.43064345595649)--(16.524055689306913,-8.775538382249595)--cycle, linewidth(0.4) + rvwvcq); /* draw figures */ draw((-9.207161569498936,12.160844964529835)--(-13.65757092076063,-9.43064345595649), linewidth(0.4) + rvwvcq); draw((-13.65757092076063,-9.43064345595649)--(16.524055689306913,-8.775538382249595), linewidth(0.4) + rvwvcq); draw((16.524055689306913,-8.775538382249595)--(-9.207161569498936,12.160844964529835), linewidth(0.4) + rvwvcq); draw(circle((1.262822751839856,-1.2516149962838792), 17.01513014612322), linewidth(0.4) + dtsfsf); draw(circle((-9.03674193706565,4.3093690417106725), 7.8533252204227315), linewidth(0.4) + linetype("2 2") + wrwrwr); draw((-9.207161569498936,12.160844964529835)--(-8.615328719291325,-15.105739920034928), linewidth(0.4) + wrwrwr); draw((-15.69490840146251,0.1448118803658121)--(12.818858446805242,11.2372877940415), linewidth(0.4) + wrwrwr); draw((6.237026976644165,15.02019947596998)--(-8.615328719291325,-15.105739920034928), linewidth(0.4) + wrwrwr); draw((-1.3146476639940998,5.739046297430744)--(-12.298037341938633,-2.8347662663502065), linewidth(0.4) + wrwrwr); draw((12.818858446805242,11.2372877940415)--(-13.65757092076063,-9.43064345595649), linewidth(0.4) + linetype("2 2") + wvvxds); draw((-1.3146476639940998,5.739046297430744)--(-13.65757092076063,-9.43064345595649), linewidth(0.4) + wrwrwr); draw((-12.298037341938633,-2.8347662663502065)--(16.524055689306913,-8.775538382249595), linewidth(0.4) + wrwrwr); draw((2.1875170149572605,15.738370124641593)--(-8.615328719291325,-15.105739920034928), linewidth(0.4) + wrwrwr); draw((6.237026976644165,15.02019947596998)--(-1.3146476639940998,5.739046297430744), linewidth(0.4) + wrwrwr); draw((-13.65757092076063,-9.43064345595649)--(2.1875170149572605,15.738370124641593), linewidth(0.4) + linetype("2 2") + wrwrwr); draw((-15.69490840146251,0.1448118803658121)--(-9.207161569498936,12.160844964529835), linewidth(0.4) + wrwrwr); /* dots and labels */ dot((-9.207161569498936,12.160844964529835),linewidth(4pt) + dotstyle); label("$A$", (-10.381878491140686,12.655462615747414), NE * labelscalefactor); dot((-13.65757092076063,-9.43064345595649),linewidth(4pt) + dotstyle); label("$B$", (-15.080746177707685,-10.900703023489811), NE * labelscalefactor); dot((16.524055689306913,-8.775538382249595),linewidth(4pt) + dotstyle); label("$C$", (17.007573945032746,-10.035122133859046), NE * labelscalefactor); dot((-8.866322304632368,-3.5421068811084915),linewidth(4pt) + dotstyle); label("$H$", (-10.691014523151672,-4.408846351259079), NE * labelscalefactor); dot((-1.3146476639940998,5.739046297430744),linewidth(4pt) + dotstyle); label("$E$", (-2.1588600396484363,6.967359626745249), NE * labelscalefactor); dot((-12.298037341938633,-2.8347662663502065),linewidth(4pt) + dotstyle); label("$F$", (-13.596893224054948,-3.8524014936393027), NE * labelscalefactor); dot((-8.615328719291325,-15.105739920034928),linewidth(4pt) + dotstyle); label("$P$", (-9.207161569498936,-16.897942044502962), NE * labelscalefactor); dot((-15.69490840146251,0.1448118803658121),linewidth(4pt) + dotstyle); label("$T$", (-17.059216782578,-0.5137323479206409), NE * labelscalefactor); dot((12.818858446805242,11.2372877940415),linewidth(4pt) + dotstyle); label("$D$", (13.050632735292115,11.728054519714451), NE * labelscalefactor); dot((6.237026976644165,15.02019947596998),linewidth(4pt) + dotstyle); label("$Y$", (6.496948856659194,15.499514110248496), NE * labelscalefactor); dot((2.1875170149572605,15.738370124641593),linewidth(4pt) + dotstyle); label("$Q$", (2.4163532341141685,16.241440587074866), NE * labelscalefactor); dot((-6.806342502966366,1.4521400155402702),linewidth(4pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] How did you get that $(B,C;P,T)$ is harmonic?

Let $D$ be the foot from $A$ to $BC$, $M=\overline{BQ} \cap \overline{EF}$, $E'=\overline{BE} \cap \Gamma \neq B$, $P_{\infty}$ be the point at infinity on $\overline{EF}$, $K \in \Gamma$ be such that $\overline{BK} \parallel \overline{EF}$, $R=\overline{KE} \cap (AFE)$, and $S=\overline{EF} \cap \overline{BC}$. We first claim that $R$ is the A-Queue point of $ABC$. This follows directly by angle chase: \[\measuredangle REF=\measuredangle RKB=\measuredangle RAB=\measuredangle RAF,\]so $AEFR$ is concyclic. Thus, $\overline{AR} \cap \overline{BC}=S$. The rest is projection: \[(P_{\infty},M;E,F)\stackrel{B}=(K,Q;E',A)\stackrel{E}=(R,P;B,C)\stackrel{A}=(S,D;B,C)=-1,\]and the Midpoints and Parallels Lemma finishes. $\blacksquare$ [asy][asy] import geometry; import olympiad; size(8cm,0); point A = (1,3.5); point B = (0,0); point C = (5,0); point H = orthocenter(A,B,C); point D = foot(A,B,C); point EE = foot(B,A,C); point F = foot(C,A,B); circle w1 = circumcircle(A,B,C); circle w2 = circumcircle(A,EE,F); point R = intersectionpoints(w1,w2)[0]; line AH = line(A,H); point P = intersectionpoints(w1,AH)[0]; line PE = line(P,EE); point Q = intersectionpoints(w1,PE)[1]; line BQ = line(B,Q); line EF = line(EE,F); point M = intersectionpoint(BQ,EF); line RE = line(R,EE); point K = intersectionpoints(w1,RE)[1]; line BC = line(B,C); point SS = intersectionpoint(EF,BC); line BE = line(B,EE); point E1 = intersectionpoints(w1,BE)[1]; draw(w1); draw(w2); draw(A--B--C--cycle); draw(A--P); draw(A--P); draw(EE--F); draw(B--Q); draw(P--Q); draw(B--K,dashed); draw(K--R,dashed); draw(F--SS--B,dashed); draw(R--SS,dashed); dot(EE); dot(F); dot("$A$",A,N); dot("$B$",B,SW); dot("$C$",C,SE); dot("$D$",D,SW); label("$E$",EE+(0.05,0.05),NE); label("$F$",F-(0.15,0),W); dot("$P$",P,SW); dot("$Q$",Q,N); dot("$R$",R,W); dot("$K$",K,E); dot("$S$",SS,SW); dot("$M$",M,SE); dot("$E'$",E1,NE); [/asy][/asy]

Let $EF \cap BQ = M$. Now its enough to show that $(E,F;M,P_\infty) = -1$. Let $BK \parallel EF$, where $K \in \Gamma$. Let $KE \cap \Gamma = Y$ and $BE \cap \Gamma = Z$. Now we project: $(E,F;M,P_\infty)\stackrel{B}{=}(Z,A;Q,K)\stackrel{E}{=}(B,C;P,Y) = -1$. Now notice that $\angle YAB = \angle YKB = \angle YEF$ $\Rightarrow$ $\angle YAF = \angle YEF$ $\Rightarrow$ YAEF is cyclic. Now we have three circles - (AYFE), (FEBC) and (ABC). Using the radical axis theorem we get that FE, BC and AY are concurrent. Let them be concurrent in L. By Ceva we get that $(B,C;D,L) = -1$. Now projecting we get $(B,C;P,Y)\stackrel{A}{=}(B,C;D,L) = -1$ $\Rightarrow$ $(E,F;M,P_\infty) = (B,C;P,Y) = -1$ $\Rightarrow$ we are ready.

We start by defining some points. Let $D = AH \cap BC$, let $R = EF\cap BC$, let $S=(AEHF)\cap \Gamma$, distinct from $A$, let $T = BE \cap \Gamma$, distinct from $B$, let $U=SE\cap \Gamma$, distinct from $S$, let $X=BQ\cap EF$, and let $\infty$ be the point at infinity on line $EF$. Note that $R,S,A$ are collinear since $AS$ is the radical axis of $(AEFH)$ and $\Gamma$, while $R$ is the radical center of $\Gamma$, $(AEFH)$, and $(BCEF)$. Since $\angle ABU = \angle ASU = \angle ASE = \angle AFE$, we have that $FE\parallel BU$. This gives us enough information to finish with cross ratios. By Ceva-Menelaus, $(RD;BC)$ is a harmonic bundle. Thus we have: \[-1 = (RD;BC) \overset{A}{=} (SP; BC) \overset{E}{=} (UQ;TA) \overset{B}{=} (\infty X; EF),\]which tells us that $X$ must be the midpoint of $EF$, as desired.

Let $X$ be the $A$-queue point; it is well-known that $(B,C;P,X) = -1.$ Also let $E' = BE \cap \Gamma$ and $B'$ be on $\Gamma$ such that $BB' \parallel EF.$ If we let $XE$ intersect $\Gamma$ again at point $B'',$ we see that $\measuredangle XEF = \measuredangle XAF = \measuredangle XAB = \measuredangle XBB',$ so $BB'' \parallel EF$ and so $B'' = B'.$ Therefore, $X,E,B'$ are collinear. To finish, we have that $$(F,E;FE \cap BQ, \infty) \stackrel{B}{=} (A,E'; Q, B') \stackrel{E}{=} (C,B;P,X) = -1,$$so $BQ$ bisects $EF.$

Let $M = \overline{BQ} \cap \overline{EF}$. We wish to show that $M$ is the midpoint of $\overline{EF}$. Suppose $K \neq B$ is the point on $\Gamma$ such that $\overline{BK} \parallel \overline{EF}$ and denote $L \neq K = \overline{KL} \cap \Gamma$. Claim: $AEFL$ is cyclic. Proof: We proceed with angle chasing: \[\angle LAF = \angle LKB = \angle LEF. \ \square\] Note that Queue point properties give that $\overline{AR}$, $\overline{EF}$, and $\overline{BC}$ are concurrent at a point $X$. Let $D = \overline{AH} \cap \overline{BC}$ and extend $\overline{BE}$ to intersect $\Gamma$ at point $G$. Finally, EGMO 9.11 gives \[-1 = (X,D;B,C) \overset{A}{=} (L,P;B,C)\overset{E}{=} (K,Q;G,A) \overset{B}{=} (\infty,M,E,F),\] which is equivalent to the desired conclusion due to EGMO 9.8. $\square$

Let the line through $B$ parallel to $EF$ meet $\Gamma$ again at $X$. Let $BE$ meet $\Gamma$ again at $R$. Let $BQ$ meet $EF$ at $M$. We want to prove that $(FE;M\infty_{EF})=-1$. Project through $B$ to get that we want to prove $(AR;QX)=-1$. Project through $E$ to get that we want to prove $(C,B;P,EX\cap\Gamma)=-1$. Let $G$ be the queue point. It is well known that $(CB;PG)=-1$ (project through $A$). Thus we want to prove that $E,X,G$ are collinear. This is just angle chasing: \[\angle EQA=\angle EHA=\angle ACB=\angle ABX=\angle AQX\;\blacksquare\]

That's from seongbin's girlfriend sojung.

Purely projective solution. Iirc you can also do by angle chasing by adding the point at $(ABC)$ where the parallel line to $\overline{EF}$ through $B$ passes through and proving that that point lies on the line $\overline{Q_AE}$ where $Q_A$ is the $A$-Queue point, and then doing some harmonic bundle chasing. OK for now the actual solution. Redefine $H_A$ as $\overline{AH} \cap (ABC)$. Also redefine $P$ and $Q$ as intersection points of $\overline{H_AE}$ and $\overline{H_AF}$ with $(ABC)$. Also define some new points. $\bullet$ Let $K$ be the point on $(ABC)$ such that $(A,K;B,C)=-1$. $\bullet$ Let $X_A$ be the $A$-Ex point of $\triangle ABC$. $\bullet$ Let $N$ be midpoint of $\overline{EF}$. First remember that $X_A$, $H_A$, $K$ are collinear since $-1=(A,K;B,C) \overset {H_A}= (D,\overline{H_AK} \cap \overline{BC};B,C)$. Now apply DDIT from $H_A$ to $BFEC$ to get $(\overline{H_AB},\overline{H_AE})$, $(\overline{H_AC},\overline{H_AF})$; $(\overline{H_AA},\overline{H_AX_A})$ are pairs under involution. Now project from $H_A$ to $(ABC)$ to get that $(B,Q)$; $(C,P)$; $(A,K)$ are pairs under involution which mean they must concur by PoP. Now apply Pascal on $ABQH_APC$ to get that $E$, $F$, $\overline{BQ} \cap \overline{PC}$ are collinear. Now $\overline{AK} \cap \overline{EF}=N$ as $-1=(E,F;N,\infty) \overset{A}= (C,B;\overline{AN} \cap (ABC),A)$.

Pretty quick and standard. Denote by $Q_A$ the $A-$Queue Point of $\triangle ABC$. Let $X$ denote the second intersection of line $\overline{Q_AE}$ with $(ABC)$. We start off with a small observation. Claim : Lines $EF$ and $BX$ are parallel. Proof : Note that, \[\measuredangle BXQ_A = \measuredangle BAQ_A = \measuredangle FAQ_A = \measuredangle FEQ_A\]which implies the claim. Now, denote by $S$ the second intersection of the $B-$altitude with $(ABC)$, and $M$ the intersection of $BQ$ with $EF$. Then, \[-1=(BC;PQ_A) \overset{E}{=}(SA;QX) \overset{B}{=}(EF;MP_\infty)\]which implies that $M$ is indeed the midpoint of segment $EF$, and we are done.

Projecting through $B$ from $EF$ to $\Gamma$ and then through $E$ from $\Gamma$ to $\Gamma$, it suffices to prove that if $G\in \Gamma$ is such that $BG\parallel EF$ and $J\in \Gamma$ is such that $J\neq G$ and $G,E,J$ are collinear then $BPCJ$ is harmonic, for which it suffices to prove that $AJFE$ is cyclic. This is clear because$$\measuredangle AJE=\measuredangle AJG=\measuredangle ABG=\measuredangle AFE.$$Done.

Nice question

Very nice question Claim: $\color{blue}{BQ}$ bisects segment $\color{blue}{\overline{EF}}$. Proof: Toss the figure on the complex plane with $(ABC)$ as the unit circle. Assign $A,B,C,Q$ the complex numbers $a,b,c,q$ respectively. Note that, $$\boxed{h=(a+b+c),e=\frac{1}{2}\left(a+b+c-\frac{ac}{b}\right),f=\frac{1}{2}\left(a+b+c-\frac{ab}{c}\right),p=\frac{-bc}{a}}$$Let $M$ be the midpoint of $\overline{EF}$ then, $m=\frac{e+f}{2}=\frac{1}{4}\left(2(a+b+c)-\frac{ab}{c}-\frac{ac}{b}\right)=\frac{1}{2}\left((a+b+c)-\frac{a}{2}\left(\frac{b}{c}+\frac{c}{b}\right)\right)$. It is well known that if a point $P$ lies on a chord $AB$ of the unit circle then, $p+ab\overline{p}=a+b$. Thus, $$E \in PQ \iff \frac{1}{2}\left(a+b+c-\frac{ac}{b}\right)-\frac{qbc}{a}\left(\frac{1}{2}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{b}{ac}\right)=q-\frac{bc}{a}$$$$ \iff \frac{1}{2}\left(a+b+c-\frac{ac}{b}\right)+\frac{bc}{a}=q+\frac{q}{2}\left(\frac{bc}{a^2}+\frac{c}{a}+\frac{b}{a}-\frac{b^2}{a^2}\right)=q\left(1+\frac{ab+bc+ca-b^2}{2a^2}\right)$$$$ \iff q=a^2\left(\frac{a+b+c+c\left(\frac{2b}{a}-\frac{a}{b}\right)}{2a^2-b^2+ab+bc+ca}\right)=\frac{a(a^{2}b+ab^{2}+abc+2b^{2}c-a^{2}c)}{b(2a^2-b^2+ab+bc+ca)}=\frac{a\cancel{(a+b)}(ab+2bc-ca)}{b\cancel{(a+b)}(2a-b+c)}$$$$ \iff \boxed{q=\frac{a(ab+2bc-ca)}{b(2a-b+c)}}$$We wish to show that $B,M,Q$ are collinear. Thus, $$ M \in BQ \iff \frac{1}{2}\left((a+b+c)-\frac{a}{2}\left(\frac{b}{c}+\frac{c}{b}\right)\right)+\frac{bq}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{2a}\left(\frac{b}{c}+\frac{c}{b}\right)\right)=b+q$$$$ \iff \frac{1}{2}\left((a+b+c)-\frac{a}{2}\left(\frac{b}{c}+\frac{c}{b}\right)\right)-b=q\left(1+\frac{b}{2}\left(\frac{1}{2a}\left(\frac{b}{c}+\frac{c}{b}\right)-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right)\right)$$$$ \iff \frac{1}{2}\left((a-b+c)-\frac{a}{2}\left(\frac{b}{c}+\frac{c}{b}\right)\right)=q\left(1+\frac{b^2+c^2}{4ac}-\frac{ab+bc+ca}{2ca}\right)$$$$ \iff q=\frac{ac\left(2a-2b+2c-a\left(\frac{b^2+c^2}{bc}\right)\right)}{(b^2+c^2+2ac-2ab-2bc)}$$$$ \iff q=\frac{a(2abc-2b^{2}c+2bc^2-ab^2-ac^2)}{b(b^2+c^2-2bc+2a(c-b))}=\frac{a\cancel{(c-b)}(ab+2bc-ca)}{b\cancel{(c-b)}(2a-b+c)}$$$$ \iff \boxed{q=\frac{a(ab+2bc-ca)}{b(2a-b+c)}}$$Thus $P,E,Q$ are collinear $\iff B,M,Q$ are collinear. Hence, $BQ$ bisects segment $\overline{EF}$ as desired. $\blacksquare$ ($\mathcal{QED}$)

for storage.

We write angle-chasing from ABPCQ is cyclic and altitudes. ∠BAP = ∠BCP = ∠BQP = ∠FCB = ∠FEB = α, ∠PAC = ∠PBC = ∠PQC = ∠EFC = ∠EBC = β, ∠ABQ = ∠ACQ = ∠APQ = θ. ∠QBE = 90° - α - β - θ, ∠ECF = 90° - α - β. We write the sine law on triangles BEC, BEQ, QEC with sides BE, EC, QE and we get sin(θ) / sin(90° - β) = sin(90° - α - β - θ) / sin(α). We want to write the median sine on triangle BFE. This is the same as this sine equation; then we get BQ bisects FE. We are done.