Elegant Problem, although my solution is no elegant at all. Anyway, it is not hard to calculate by complex numbers.

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My solution(easier): First, construct perpendiculars from points O and P to lines AC and AB respectively, with M, G, N, H as the corresponding midpoints. Then $\frac{\sin \angle OPH}{\sin \angle OPG}=\frac{NH}{MG}=\frac{AF}{AE}=\frac{\sin \angle AEF}{\sin \angle AFE}$. Also, $\angle OPH + \angle OPG = 180^{\circ} - \angle A = \angle AEF + \angle AFE$, hence $\angle OPH = \angle AEF$, so $\angle STA = \angle AQP$ and therefore QRST is cyclic. My social credits are at a record low now. Please someone donate me social credits T_T

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Let $G$ be the point that satisfies $\overleftrightarrow{AG} \parallel \overleftrightarrow{EF}$ and $GA = GB$. Let $H$ be the point that satisfies $\overleftrightarrow{AH} \parallel \overleftrightarrow{EF}$ and $HA = HC$. Let $I$ be the intersection of $\overleftrightarrow{EF}$ and $\overleftrightarrow{BG}$. Let $J$ be the intersection of $\overleftrightarrow{EF}$ and $\overleftrightarrow{CH}$. Notice that $\triangle BIF \sim \triangle BGA$. We have $IB = IF$. Then $\overleftrightarrow{PI}$ is the perpendicular bisector of $\overline{BF}$ and hence $\overleftrightarrow{PI} \parallel \overleftrightarrow{OG}$. Similarly we have $\overleftrightarrow{PJ} \parallel \overleftrightarrow{OH}$. From $\overleftrightarrow{PI} \parallel \overleftrightarrow{OG}$, $\overleftrightarrow{PJ} \parallel \overleftrightarrow{OH}$ and $\overleftrightarrow{IJ} \parallel \overleftrightarrow{GH}$, we have $\overleftrightarrow{GI}$, $\overleftrightarrow{HJ}$ and $\overleftrightarrow{OP}$ are concurrent at a point $K$. $\measuredangle BKC = \overleftrightarrow{CK} - \overleftrightarrow{BK} = (2\overleftrightarrow{CE} - \overleftrightarrow{EJ}) - (2\overleftrightarrow{BF} - \overleftrightarrow{FI}) = 2\measuredangle BAC = \measuredangle BOC$, so $B$, $O$, $C$ and $K$ are concyclic. Thus, $\overleftrightarrow{OP} = \overleftrightarrow{BO} - \overleftrightarrow{BC} + \overleftrightarrow{CK} = 90^\circ - \measuredangle BAC + 2\overleftrightarrow{AC} - \overleftrightarrow{EF} = 90^\circ - \overleftrightarrow{EF} + \overleftrightarrow{AC} + \overleftrightarrow{AB} = \overleftrightarrow{AC} + \overleftrightarrow{AB} - \overleftrightarrow{ST} \implies Q$, $R$, $S$ and $T$ are concyclic.

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Let $A_\circ^*$, $A_\bullet^*$ be the antipodes of $A$ with respect to $\odot(ABC)$, $\odot(AEF)$ respectively. Then $P$ is the midpoint of $\overline{A_\circ^*A_\bullet^*}$, and hence \[ST = OP = AA_\bullet^* = FA + AE - EF + 90^\circ = RT + SQ - QR. \]

Nice problem Let $\omega_1=(AEF)$, $\omega_2=(ABC)$. $\omega_1 \cap \omega_2 = M \neq A$. The circle with diameter $AP$ passes through $A$ and the midpoints of $BF$ and $EC$, so it is the locus of points $X$ such that $\text{Pow}(X,\omega_1)+\text{Pow}(X,\omega_2)=0$. Then $\angle AMP = 90^\circ$, so $P$ is the midpoint of the antipodes $S_1, S_2$ of $A$ in $\omega_1, \omega_2$ respectively. Hence $OP \parallel AS_1$, so we have \begin{align*} &\angle OP + (\angle EF - 90^\circ) - \angle AB - \angle AC = \angle AS_1 + \angle EF - (\angle AB + 90^\circ) - \angle AC \\ &= \angle AS_1 + \angle EF - \angle FS_1 - \angle AE = 0, \end{align*}hence proved.

Let $M$ and $N$ be the midpoints of $CE$, $BF$. By forgotten coaxiality lemma as $M,N$ are midpoints then $P$ is the midpoint of the antipode of $A$ on $(AEF)$ (called $H$) and $(ABC)$. Then, by midpoints $AH \parallel PO$. To finish we angle chase $\angle TRQ=\angle FAH=\angle FEH=\angle PSE$ from the right angles

Solution from Twitch Solves ISL: Note that \begin{align*} \measuredangle QST &= \measuredangle(\overline{PST}, \overline{AC}) = 90^{\circ} + \measuredangle(\overline{EF}, \overline{AC}) \\ \measuredangle QRT &= \measuredangle(\overline{AB}, \overline{OP}). \end{align*}So we can erase all the points $Q$, $R$, $S$, $T$ from the picture and focus just on proving the claim that: Claim: $\measuredangle(\overline{AB}, \overline{OP}) = \measuredangle(\overline{EF}, \overline{AC}) + 90^{\circ}$. [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ pair A = (-2.,2.); pair B = (-4.,-2.5); pair C = (2.5,-2.5); pair F = (-3.45808,-1.28069); pair E = (0.33045,-0.33045); pair P = (-0.49655,-3.32700); pair O = (-0.75,-1.25); pair Q = (-1.02796,1.02796); pair R = (-1.33041,3.50656); pair S = (-1.77664,1.77664); pair T = (-1.89303,2.24067); pair M = (1.41522,-1.41522); pair N = (-3.72904,-1.89034); import graph; size(9.14447cm); pen zzttqq = rgb(0.6,0.2,0.); pen cqcqcq = rgb(0.75294,0.75294,0.75294); 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(O, 3.48209), linewidth(0.6)); draw(P--R, linewidth(0.6)); draw(A--R, linewidth(0.6)); draw(P--T, linewidth(0.6)); draw(E--F, linewidth(0.6)); draw(P--M, linewidth(0.6)); draw(P--N, linewidth(0.6)); dot("$A$", A, dir((-25.918, 4.569))); dot("$B$", B, dir((-27.115, -23.952))); dot("$C$", C, dir((4.415, -24.858))); dot("$F$", F, dir((-22.398, 17.250))); dot("$E$", E, dir((3.860, 9.229))); dot("$P$", P, dir((-9.506, -31.881))); dot("$O$", O, dir((4.057, 6.930))); dot("$Q$", Q, dir((3.758, 7.519))); dot("$R$", R, dir((3.189, 7.076))); dot("$S$", S, dir((-16.534, -26.566))); dot("$T$", T, dir((-7.614, 24.004))); dot("$M$", M, dir((3.232, 7.139))); dot("$N$", N, dir((-15.240, 19.306))); [/asy][/asy] Proof. Use complex numbers with $ABC$ the unit circle. Let $P$ be any point, with coordinate $p$. Then the foot from $P$ to $\overline{AC}$ has coordinates \[ M \coloneqq \frac{1}{2} \left( p + a + c - ac \overline p \right) \]and so (since $M$ is the midpoint of $\overline{EC}$) we get \[ E = 2 \cdot M - c = p + a - ac \overline p. \]Analogously \[ F = p + a - ab \overline p. \]So then we have the ratio \[ z \coloneqq \frac{E-F}{a-c} \div \frac{a-b}{p-0} = \frac{a\overline p(b-c) \cdot p}{(a-b)(a-c)} = p \bar p \cdot \frac{a(b-c)}{(a-b)(a-c)}. \]The complex conjugate is \[ \bar p p \cdot \frac{\frac 1a\left( \frac 1b- \frac 1c \right)} {\left( \frac 1a - \frac 1b \right)\left( \frac 1a - \frac 1c \right)} = -z. \]so $z$ is pure imaginary and we're done. $\blacksquare$

Let the perpendicular bisectors of $CE$ and $BF$ meet line $EF$ at $Y$ and $Z$, and let lines $BZ$ and $CY$ intersect at $X$. Then $P$ is the incenter of $\triangle XYZ$, and if we let $O'$ be the second intersection of $(BXC)$ and line $XP$, then $BO' = O'C$ and $\angle BOC = 180^\circ - \angle X = \angle Y + \angle Z = 2\angle A$, so $O' = O$, and hence $X$, $P$, $O$ are collinear. We then have \[ \measuredangle PTF = 90^\circ - \measuredangle TFE = 90^\circ - \measuredangle BFZ = \measuredangle FZP = \measuredangle EQP = \measuredangle SQR, \]so $Q$, $R$, $S$, $T$ are concyclic.

Construct $A_1$ and $A_2$ as the antipodes of $A$ with respect to $(AEF)$ and $(ABC)$. It is well known that $P$ is the midpoint of $A_1A_2$. Let $O_1$ and $H_1$ be the circumcenter and orthocenter of $(AEF)$. Then notice that $PO\parallel AA_2$. Thus since $AO_1$ and $AH_1$ are anti-parallel with respect to $\angle BAC$, then $PO$ and the perpendicular through $P$ to $EF$ are also anti-parallel with respect to $\angle BAC$.

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Is this a valid solution to the problem? Please let me know

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