I'm told that this problem was proposed by Brandon Wang and Eric Gan. I can't confirm whether that's true or not, though.

This problem was proposed by Ankan Bhattacharya.

v_Enhance wrote: This problem was proposed by Ankan Bhattacharya. Oh, thanks for the clarification. Solution with Brandon Wang, Eric Gan, and 0 others: Let $Q^* = \overline{BE} \cap \overline{CF}$. Then \[\angle BCQ^* = \angle DCF = \angle DAF = \angle PAB = \angle PCB = \angle BCQ\]and analogously $\angle CBQ^* = \angle CBQ$, where we direct angles mod $180^\circ$. It follows that $Q = Q^*$. Finally, to show that $Q$ lies on the circumcircle of $\triangle AEF$, note that \[\angle AFQ = \angle AFC = \angle ADC = \angle ADB = \angle AEB = \angle AEQ\]and we are done.

Moving points solution: Animate $D$ linearly on $BC$. Then $D \mapsto P \mapsto Q$ is a projective map (last map from reflection in $BC$). Also note that $Q$ moves on $\odot (BHC)$, where $H$ is orthocenter of $\triangle ABC$. Now consider $\sqrt{bc}$ inversion (denoting inverse of object $Z$ with $Z'$). Then $D \mapsto D' \mapsto E'$ is projective (last map by perspectivity at $B$ from $\odot (ABC)$ to $AC$). This gives $\text{deg}(E')=1$. Similarly, we get $\text{deg}(F')=1$. Also, $\text{deg}(Q')=2$ since $Q'$ moves projectively on a circle. Thus, it suffices to show that $E',F',Q'$ are collinear (i.e. prove the original problem) for $(1+1+2)+1=5$ positions of $D$. $D=B,C$ are obvious. Take $D$ as the foot of the $A$-altitude for the third position (and use the fact that reflection of orthocenter in $BC$ lies on circumcircle). Then take $D$ as the foot of the $A$-tangent. Then $\sqrt{bc}$ inversion gives that $E'$ and $F'$ lie on the perpendicular bisector of $BC$, and $Q'$ is the circumcenter of $\triangle ABC$. These are collinear, as desired. Finally consider $D$ as the foot of the $A$-symmedian. Then $Q$ is well known to be the $A$-Humpty point. In this case, the result follows from here. Hence, done. $\blacksquare$

By Angle Chasing, we get that $$\angle BQD = \angle BPD = \angle BCA = \angle BED$$So, $BFDQ$ is conncyclic. Similarly, we get that $CEQD$ is concyclic. Therefore \begin{align*} \angle EQF &= 360^\circ - \angle DQF - \angle DQE \\ &= 360^\circ - (180^\circ - \angle CBA) - (180^\circ - \angle QCB) \\ &= 180^\circ - \angle BAC \end{align*}which implies that $Q$ lies on the circumcircle of $\triangle AEF$

Note that we have ... $$\angle FCB = \angle FAD = \angle FAP = \angle BAP=\angle BCP = \angle BCQ$$So , $Q$ lies on $CF$ $$\angle CFA = \angle ADC = \angle BDP= \angle BDQ \implies BDQF \text{ is cyclic} $$Similarly , $CEQD$ is cyclic... We are done by Miquel on ${{(D,E,F)}}$

alifenix- wrote: Let $ABC$ be a triangle. Distinct points $D$, $E$, $F$ lie on sides $BC$, $AC$, and $AB$, respectively, such that quadrilaterals $ABDE$ and $ACDF$ are cyclic. Line $AD$ meets the circumcircle of $\triangle ABC$ again at $P$. Let $Q$ denote the reflection of $P$ across $BC$. Show that $Q$ lies on the circumcircle of $\triangle AEF$. Let $Q'=BE\cap CF$.We will first show that $Q'\in \odot{AEF}$.Indeed since $$\measuredangle{FQ'E}=-\measuredangle{BQ'C}=\measuredangle{Q'CB}+\measuredangle{CBQ'}=\measuredangle{FCB}+\measuredangle{CBE}=\measuredangle{FCD}+\measuredangle{DBE}=\measuredangle{FAD}+\measuredangle{DAE}=\measuredangle{FAE}$$which proves this Claim$\square$. Back to the main problem we have $\measuredangle{CPD}=\measuredangle{CBA}=-\measuredangle{CED}=-\measuredangle{CQD}$ and similarly $\measuredangle{BPD}=-\measuredangle{BQD}$. hence $Q'$ is reflection of $P$ about $BC$ which implies $Q'\equiv Q$ and we are done$\square$

It's quite interesting that both EGMO TST geometry problems were basically extensions of ELMO 2013 G3.

Solution: We use $\sqrt{bc}$ inversion. $D \mapsto D'$ on $(ABC)$ such that $\angle{BAP} = \angle{D'AC}$; $F \mapsto F' = BD' \cap AC$, $E \mapsto E' = CD' \cap AB$. We identify $Q'$ as follows: let $A_1$ be the reflection of $A$ in $BC$; then $Q, D, A_1$ are collinear and $QBA_1C$ is cyclic. Let $O$ be the circumcenter of $ABC$; we know that $A_1$ and $O$ swap under this inversion. Then $O, D', A,$ and $Q'$ must be concyclic; so must $B, O, C,$ and $Q'$. Thus $Q'$, which we want to show lies on $E'F'$, is the intersection of $(BOC)$ and $(AOD')$. But then $Q'$ is just the Miquel point of $ABD'C$ so we are done. @above haha, solved both the same way

alifenix- wrote: Let $ABC$ be a triangle. Distinct points $D$, $E$, $F$ lie on sides $BC$, $AC$, and $AB$, respectively, such that quadrilaterals $ABDE$ and $ACDF$ are cyclic. Line $AD$ meets the circumcircle of $\triangle ABC$ again at $P$. Let $Q$ denote the reflection of $P$ across $BC$. Show that $Q$ lies on the circumcircle of $\triangle AEF$. The problem is quite trivial for any TST . Anyways, here goes the solution: Let $BE$ meet $CF$ at point $T$. We first prove the following. Claim.The points $A,E,F,T$ are concyclic. Proof. This simply follows because \begin{align*}\measuredangle{ETF} & =-\measuredangle{CTB}\\& =\measuredangle{BCT}+\measuredangle{TBC}\\& =\measuredangle{BCF}+\measuredangle{CBE}\\& =\measuredangle{FCD}+\measuredangle{DBE}\\& =\measuredangle{FAD}+\measuredangle{DAE}\\& =\measuredangle{EAF}. ~~~~~~~~~~~~~\square\end{align*} Therefore, we have $$\measuredangle{DPC}=\measuredangle{ABC}=-\measuredangle{DEC}=-\measuredangle{DQC}$$and similarly $\measuredangle{DPB}=-\measuredangle{DQB}$. Hence, we obtain that $T$ is reflection of $P$ about $BC$, which implies $Q\equiv T$ and hence, the conclusion follows.$\blacksquare$

Nice , but quite trivial for a #4. Claim. $BDQF$ is cyclic. Proof. Notice that \[ \measuredangle BFD = \measuredangle BCA = \measuredangle BPA \equiv \measuredangle BPD = \measuredangle BQD \] Similarly, $DQEC$ is cyclic. Hence, by Miquel Theorem, $AQFE$ is cyclic as well.

wu2481632 wrote: Solution: We use $\sqrt{bc}$ inversion. $D \mapsto D'$ on $(ABC)$ such that $\angle{BAP} = \angle{D'AC}$; $F \mapsto F' = BD' \cap AC$, $E \mapsto E' = CD' \cap AB$. We identify $Q'$ as follows: let $A_1$ be the reflection of $A$ in $BC$; then $Q, D, A_1$ are collinear and $QBA_1C$ is cyclic. Let $O$ be the circumcenter of $ABC$; we know that $A_1$ and $O$ swap under this inversion. Then $O, D', A,$ and $Q'$ must be concyclic; so must $B, O, C,$ and $Q'$. After this we can also proceed in the following way: Let $X$ be the intersection of $AD'$ and $BC$.Then by brocards theorem $E'F'$ is the polar of $X$. Again $Q'$ is the intersection of the circumcircle of $(BOC)$ and $OD'$.So inversion in $(ABC)$ sends $X$ to $Q'$.So $Q'$ also lies in polar of $X$.So $E',Q',F'$ are collinear.So inverting back $A,E,Q,F$ lies on a circle.

Let $(DEF)$ meet $BC$ again at $G$, and $T$ be the point such that $\triangle BAQ$ and $\triangle TAC$ are directly similar. Then we have $GE\parallel CT$ and $GF\parallel BT$, so this problem becomes a corollary of problem 2 of the same TST. Unfortunately, proving these claims might require solving this problem first.

Applying inversion of center $A $, and an arbitary $\mathrm {power} $. Converting the problem into this lemma. Lemma- Let $O $ be the circumcircle of $\triangle ABC $, $D\in\odot (O) $. Let $AD $ cuts $BC $ and $BD$ cuts $AC $, $CD $ cuts $AB $ at $P $, $E $, $F $ respectively. Let $\omega $ be a circle passes through $A $, $P $ such that $(O) $ and $\omega $ are orthogonal. $\omega $, $OP $, $EF $ are concurrent at $Q $. $Q $ is the Miquel point of the complete quadrilateral $ABCD\cdot EF.$ Proof- Let $Q_1$ be the Miquel point. By Brocard's theorem $OP\cdot OQ_1=OA^2\implies OA $ touches $\odot (APQ_1)$. $\mathrm {Pow (O)} $ wrt $\odot (APQ_1)=OA^2$. $(O) $ and $(APQ_1) $ both are orthogonal $\implies Q=Q_1$.

USATST for EGMO 2020 P4 wrote: Let $ABC$ be a triangle. Distinct points $D$, $E$, $F$ lie on sides $BC$, $AC$, and $AB$, respectively, such that quadrilaterals $ABDE$ and $ACDF$ are cyclic. Line $AD$ meets the circumcircle of $\triangle ABC$ again at $P$. Let $Q$ denote the reflection of $P$ across $BC$. Show that $Q$ lies on the circumcircle of $\triangle AEF$. Proposed by Ankan Bhattacharya Let $\{\omega_1,\omega_2\}=\{\odot(ABDE),\odot(AFDC)\}$ respectively. Consider circles $\omega_1,\omega_2$. Notice that the Spiral Similarity centered at $D$, $\triangle DFB\mapsto DCE$. So, $\angle FDB=\angle EDC$ and as $Q$ is the reflection of $P$ on $BC$ we get that $\angle ADB=\angle QDC\implies \angle ADF=\angle QDE$. Now $\angle QBD=\angle CBP=\angle EAD=\angle EBD\implies \overline{B-Q-E}$. Similarly we get that $\overline{C-Q-F}$.So, $\angle BQD=\angle BFD=\angle ECD\implies Q\in\odot(BFD)$ and $Q\in\odot(CED)$. So, $Q$ is the Miquel Point of $ABC$ WRT $\triangle DEF\implies Q\in\odot(AEF)$. $\blacksquare$

[asy][asy] import geometry; size(4cm); defaultpen(fontsize(10pt)); unitsize(1.5cm); point A=(0.5,3); point B=(0,0); point C=(4,0); point D=(B+C)/2.5; circle O1=circle(A,B,D); circle O2=circle(A,C,D); circle cABC=circle(A,B,C); point E_=intersectionpoints(O1, line(A,C))[0]; point F=intersectionpoints(O2, line(A,B))[0]; point P=intersectionpoints(cABC, line(A,D))[0]; point Q=reflect(B,C)*P ; point M=(P+Q)/2 ; draw(A--B--C-- cycle ); draw(O1 ^^ O2 ^^ cABC); draw(B--Q--C ^^ B--P--C); draw(F--Q--E_, dashed); draw(A--P ^^ P--Q , gray); markrightangle(Q,M,C, gray, size=0.2cm); markangle(C,B,Q, radius=0.5cm); markangle(P,B,C, radius=0.6cm); markangle(P,A,C, radius=0.5cm, gray); dot("$A$", A, N+W); dot("$B$", B, S+W); dot("$C$", C, E+S); dot("$D$", D, 2*S+W); dot("$E$", E_, E+N); dot("$F$", F, 2*W+N); dot("$P$", P, S); dot("$Q$", Q, N); [/asy][/asy] Since we have $Q$ is the reflection of $P$ across $BC$, then \begin{align*} \angle DBQ &=\angle DBP=\angle PBC\\ &=\angle PAC=\angle DAE\\ &=\angle DBE \end{align*}Therefore, the points $B,Q,E$ are collinear. Similarly, the points $C,Q,F$ are also collinear. Hence \begin{align*} \angle FQE &=\angle BQC=\angle BPC\\ &=\pi-\angle A \end{align*}Finally, this implies that $AEQF$ is concyclic.

SivmengMaths wrote: [asy][asy] import geometry; size(4cm); defaultpen(fontsize(10pt)); unitsize(1.5cm); point A=(0.5,3); point B=(0,0); point C=(4,0); point D=(B+C)/2.5; circle O1=circle(A,B,D); circle O2=circle(A,C,D); circle cABC=circle(A,B,C); point E_=intersectionpoints(O1, line(A,C))[0]; point F=intersectionpoints(O2, line(A,B))[0]; point P=intersectionpoints(cABC, line(A,D))[0]; point Q=reflect(B,C)*P ; point M=(P+Q)/2 ; draw(A--B--C-- cycle ); draw(O1 ^^ O2 ^^ cABC); draw(B--Q--C ^^ B--P--C); draw(F--Q--E_, dashed); draw(A--P ^^ P--Q , gray); markrightangle(Q,M,C, gray, size=0.2cm); markangle(C,B,Q, radius=0.5cm); markangle(P,B,C, radius=0.6cm); markangle(P,A,C, radius=0.5cm, gray); dot("$A$", A, N+W); dot("$B$", B, S+W); dot("$C$", C, E+S); dot("$D$", D, 2*S+W); dot("$E$", E_, E+N); dot("$F$", F, 2*W+N); dot("$P$", P, S); dot("$Q$", Q, N); [/asy][/asy] Since we have $Q$ is the reflection of $P$ across $BC$, then \begin{align*} \angle DBQ &=\angle DBP=\angle PBC\\ &=\angle PAC=\angle DAE\\ &=\angle DBE \end{align*}Therefore, the points $B,Q,E$ are collinear. Similarly, the points $C,Q,F$ are also collinear. Hence \begin{align*} \angle FQE &=\angle BQC=\angle BPC\\ &=\pi-\angle A \end{align*}Finally, this implies that $AEQF$ is concyclic. Your Solution is similar to #7 , #9 and #18. Anyways , you presented the solution in a much simpler manner.

[asy][asy] import geometry; size(4cm); defaultpen(fontsize(10pt)); unitsize(1.5cm); point A=(0.5,3); point B=(0,0); point C=(4,0); point D=(B+C)/2.5; circle O1=circle(A,B,D); circle O2=circle(A,C,D); circle cABC=circle(A,B,C); point E_=intersectionpoints(O1, line(A,C))[0]; point F=intersectionpoints(O2, line(A,B))[0]; point P=intersectionpoints(cABC, line(A,D))[0]; point Q=reflect(B,C)*P ; point M=(P+Q)/2 ; draw(A--B--C-- cycle ); draw(O1 ^^ O2 ^^ cABC); draw(B--Q--C ^^ B--P--C); draw(F--Q--E_, dashed); draw(A--P ^^ P--Q , gray); markrightangle(Q,M,C, gray, size=0.2cm); markangle(C,B,Q, radius=0.5cm); markangle(P,B,C, radius=0.6cm); markangle(P,A,C, radius=0.5cm, gray); dot("$A$", A, N+W); dot("$B$", B, S+W); dot("$C$", C, E+S); dot("$D$", D, 2*S+W); dot("$E$", E_, E+N); dot("$F$", F, 2*W+N); dot("$P$", P, S); dot("$Q$", Q, N); [/asy][/asy] (diagram stolen from @SivmengMaths) Redefine $Q=BE\cap CF$. Now we have, \begin{align*} \measuredangle EQF&=\measuredangle BQC\\ &=\measuredangle QBC+\measuredangle BCQ\\ &=\measuredangle EBD+\measuredangle DCF\\ &=\measuredangle EAD+\measuredangle DAF\\ &=\measuredangle EAF .\end{align*}This gives that $AFQE$ is cyclic. So now we get that, \begin{align*} \measuredangle PCB=\measuredangle PAB&=\measuredangle DAF\\ &=\measuredangle DCF\\ &=\measuredangle BCQ ,\end{align*}and similarly $\measuredangle PBC=\measuredangle CBQ$, which finally gives that $P$ is the reflection of $Q$ over $BC$ (due to a kite) and we are done.

We have $\angle BQD=\angle BPD=\angle BCA=\angle BFD\Rightarrow BFQD$ is cyclic Similarly, $CEQD$ is cyclic, which means $Q$ is the Miquel point of $\triangle ABC\Rightarrow Q\in (AEF).$

We uploaded our solution https://calimath.org/pdf/USAEGMOTST2020-4.pdf on youtube https://youtu.be/D4NBqdboUbE.

Claim: $\triangle FDQ \sim \triangle ADE$. Note that \[\angle FDQ = 180^{\circ} - \angle FDB - \angle QDC = 180^{\circ} - \angle A - \angle CDP = 180^{\circ} - \angle EDC - \angle ADB = \angle ADE.\]Furthermore, \begin{align*} & \frac{FD}{DQ} \\ = ~ & \frac{FD}{DP} \\ = ~ & \frac{FD \cdot AD}{BD \cdot DC} \\ = ~ & \frac{AD}{BD \cdot \frac{DC}{FD}} \\ = ~ & \frac{AD}{DE}, \end{align*}so by SAS similarity our claim is proven. So, $\angle FQD = \angle AED = 180^{\circ} - \angle B$; similarly, $\angle EQD = 180^{\circ} - \angle C$, so \[\angle FQE = 360^{\circ} - \angle FQD - \angle EQD = 180^{\circ} - \angle A\]as desired.

good example of phantom points Let $Q'$ be the Miquel point of $\triangle DEF$ w.r.t $\triangle ABC$. Notice that \[\measuredangle BQ'D=\measuredangle BFD=\measuredangle ACB=\measuredangle DPB\]\[\measuredangle CQ'D=\measuredangle CED=\measuredangle ABC=\measuredangle DPC\]which implies $Q'$ is the reflection of $P$ over $BC$, thus $Q'\equiv Q$, done.

Note that \[\angle CBP=\angle CAP=\angle EAD=\angle EBC,\]so $Q$ lies on $BE$. Similarly, $Q$ lies on $CF$, so $Q=BE\cap CF$. Then \[\angle AFQ+\angle AEQ=\angle AFC+\angle AEB=\angle ADC+\angle ADB=180^{\circ}\;\blacksquare\]

We will prove that $Q = \overline{CF} \cap \overline{BE}$. Note that if and only if $Q$ lies on $CF$, then $\angle{BCP} =\angle{FCB}$. \[\angle BCP = \angle BAP = \angle FCD\]and then by symmetry $Q$ lies on $BE$. Now we can prove that $AEFQ$ is cyclic. \[\angle AEQ + \angle AFQ = \angle ADB + \angle ADC = 180^{\circ}\]so we are done.

deleted $\quad$

Lovely problem Here is my solution without phantom points $\textbf{Claim:}$ Points $B-E-Q$-are collinear $\textbf{Proof:}$ $\angle PBC=\angle PAC \equiv \angle PAE=\angle DBE \equiv \angle CBE \equiv \angle EBC \implies \angle PBC=\angle EBC$ $...(1)$ Since $Q$-is the reflection of $P$ over $BC$ we get: $\angle PBC=\angle QBC$ $...(2)$ Combining $(1)$ with $(2)$ we get: $\angle EBC= \angle QBC$ So Points $B-E-Q$-are collinear $\textbf{Claim:}$ Points $C-Q-F$-are collinear $\textbf{Proof:}$ $\angle QCB=\angle BCP=\angle BAP=180-\angle PAF=\angle QCB=\angle FCB \implies \angle QCB \equiv \angle FCB$ So Points $C-Q-F$-are collinear $\textbf{Claim:}$ $Q \in \odot$ $( \triangle AEF )$ $\textbf{Proof:}$ $\angle FQE=\angle FQB=180-\angle BQC \equiv 180-\angle BPC=180-(180- \angle A)=\angle A \implies \angle FQE=\angle A$ $...(*)$ $\angle A + \angle FAC=180 \implies \angle FAC= 180-\angle A , \angle FAE \equiv \angle FAC=180-\angle A \implies \angle FAE=180-\angle A$ $ ...(**)$ $(*) + (**) \implies \angle FQE+\angle FAE=\angle A + 180-\angle A=180 \implies \angle FQE+\angle FAE=180$ So Points $A,F,Q,E$-are concylic $\iff$ $Q$ lies on the circumcircle of $\triangle AEF$ $\blacksquare$

I Love Colorful Geo diagrams. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); 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 = -19.68117321949262, xmax = 22.89738128743351, ymin = -15.250512502480996, ymax = 9.540656309645739; /* image dimensions */ pen ffvvqq = rgb(1,0.3333333333333333,0); pen ffqqtt = rgb(1,0,0.2); pen ffdxqq = rgb(1,0.8431372549019608,0); pen qqccqq = rgb(0,0.8,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen wwqqcc = rgb(0.4,0,0.8); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); draw((-2.5886072783962946,6.149935911574594)--(-9.774306840329475,-7.809119271185058)--(7.045384433858,-10.037595717607374)--cycle, linewidth(1) + ffvvqq); /* draw figures */ draw((-2.5886072783962946,6.149935911574594)--(-9.774306840329475,-7.809119271185058), linewidth(1) + ffvvqq); draw((-9.774306840329475,-7.809119271185058)--(7.045384433858,-10.037595717607374), linewidth(1) + ffvvqq); draw((7.045384433858,-10.037595717607374)--(-2.5886072783962946,6.149935911574594), linewidth(1) + ffvvqq); draw(circle((-0.6681263925632243,-3.6676876138441528), 10.003698242135544), linewidth(1) + blue); draw((-2.5886072783962946,6.149935911574594)--(-0.11930894731573694,-13.656320049342908), linewidth(1) + ffqqtt); draw(circle((-4.356422343706356,-1.7690643722696051), 8.113922331441293), linewidth(1) + ffdxqq); draw(circle((4.34477874864532,-0.684262611748228), 9.735405038115452), linewidth(1) + qqccqq); draw(circle((0.6564827975021906,1.2143606298263208), 5.906819191620764), linewidth(1) + fuqqzz); draw((-9.774306840329475,-7.809119271185058)--(1.1851603191548814,-4.668751900169902), linewidth(1) + ffqqtt); draw((1.1851603191548814,-4.668751900169902)--(7.045384433858,-10.037595717607374), linewidth(1) + ffqqtt); draw((7.045384433858,-10.037595717607374)--(-0.11930894731573694,-13.656320049342908), linewidth(1) + ffqqtt); draw((-9.774306840329475,-7.809119271185058)--(-0.11930894731573694,-13.656320049342908), linewidth(1) + ffqqtt); draw((-0.11930894731573694,-13.656320049342908)--(1.1851603191548814,-4.668751900169902), linewidth(1) + wwqqcc); draw((-5.245984907664572,0.9876715212154584)--(1.1851603191548814,-4.668751900169902), linewidth(1) + ubqqys); draw((1.1851603191548814,-4.668751900169902)--(3.447153510064016,-3.991661838099557), linewidth(1) + ubqqys); /* dots and labels */ dot((-2.5886072783962946,6.149935911574594),dotstyle); label("$A$", (-2.477436117803798,6.427863813055835), NE * labelscalefactor); dot((-9.774306840329475,-7.809119271185058),dotstyle); label("$B$", (-9.675768766167941,-7.524116841302483), NE * labelscalefactor); dot((7.045384433858,-10.037595717607374),dotstyle); label("$C$", (7.166662063595267,-9.747540053152415), NE * labelscalefactor); dot((-0.11930894731573694,-13.656320049342908),dotstyle); label("$P$", (-0.0038777946207523622,-13.388395562556676), NE * labelscalefactor); dot((-0.6983774444780035,-9.011608339405127),linewidth(4pt) + dotstyle); label("$D$", (-0.5875263877313587,-8.802585188116193), NE * labelscalefactor); dot((3.447153510064016,-3.991661838099557),linewidth(4pt) + dotstyle); label("$E$", (3.5535993443391334,-3.772090171305724), NE * labelscalefactor); dot((-5.245984907664572,0.9876715212154584),linewidth(4pt) + dotstyle); label("$F$", (-5.145543972023712,1.2028192652084968), NE * labelscalefactor); dot((1.1851603191548814,-4.668751900169902),dotstyle); label("$Q$", (1.3023833423410807,-4.383531554564455), NE * labelscalefactor); label("$m$", (2.4418877384141693,-4.744837826490069), NE * labelscalefactor,ubqqys); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] $\color{red}\textbf{Claim:-}$ $AFQE$ are cyclic points. $\color{blue}\textbf{Proof:-}$ Since $Q$ is the reflection of $P$ over $DC$ therefore we get, $BP=BQ$ and $BC=CQ.$ Therefore we get some angles equal $$\angle PBD=\angle PBQ=\angle CBQ=\angle DBQ$$Now since $ABPC$ is cyclic we get, $$\angle CBP=\angle DBP=\angle DAC=\angle PAC$$Also we get, $$\angle BAD=\angle BAP=\angle BCP=\angle DCP$$Therefore we get, $B,Q,E$ are collinear and $C,Q,F$ are collinear. Now we get, $$\angle EQF=\angle BQC=\angle CPB=180-\angle BAC=180-\angle FAE.$$Q.E.D

$ $

Good problem not hard! Note that $\angle BDF = \angle A = \angle CDE$ and $\angle QDC = \angle PDC = \angle ADC$. Hence, $\angle ADF = \angle QDE$. We now claim that $\triangle DFQ \sim \triangle DAE$. indeed, $DF\cdot DE = DB \cdot \frac{AC}{AB} \cdot DC \cdot \frac{AB}{AC} = DB \cdot DC = DA \cdot DP = DA\cdot DQ$, which combined with $\angle QDF = \angle ADE$, proves our claim. We can thus get $\angle FQE + \angle FAE = 2\pi - \angle AED - \angle AFD + \angle A = \angle A + \angle FDE + \angle A = \pi$ as wanted.

Simple enough with the power of wishful thinking. Let $Q' = BE \cap CF$. We claim that $Q' = Q$. Indeed notice that $$\angle Q'BC = \angle EBD = \angle EAD = \angle CAP = \angle CBP.$$Similarly, $$\angle Q'CB = \angle PCB.$$This suffices. From here, $$\angle AFQ = \angle AFC = \angle ADC = \pi - \angle ADB = \pi - \angle AEB = \pi - \angle AFQ,$$yielding the result. $\square$

Claim 1: $B - Q - E$ collinear. Proof: $\angle DBE = \angle DAE =\angle PAC =\angle PBC =\angle DBQ$ $\square$ Similarly we prove $C - Q - F$ collinear. Claim 2: $BDQF$ cyclic Proof: $\angle QBD = \angle EBD =\angle DAC =\angle DFC =\angle DFQ$ $\square$ Similarly, we prove $DQEC$ cyclic. We are done because this means $Q$ is Miquel’s point in triangle $\triangle ABC$ for the points $D$, $E$, $F$. $\blacksquare$