Simple angle chasing. Firstly, we have $\angle EBC = \angle BEC = \angle BPC$, so $\angle PCB = 180^{\circ} - 2\angle EBC - \angle ABE$. Thus, we have $\angle PAB = 180^{\circ} - \angle PEB = 180^{\circ} - \angle PCB = 2\angle EBC + \angle ABE$. We note that $\angle PBC = \angle DBF$ and $\angle EBC = \angle DBQ$. Therefore, $\angle DFQ = \angle DBQ = \angle EBC$, and $\angle DFB = \angle DBF = \angle EBC + \angle PBE$. Therefore, $\angle QAB = 180^{\circ} - \angle QFB = 180^{\circ} - (\angle DFQ + \angle QFB) = 180^{\circ} - (2 \angle EBC + \angle PBC) = 180^{\circ} - \angle PAB$. Therefore, $\angle PAQ = 180^{\circ}$, so $A$, $P$ and $Q$ are collinear.

Can anynone please provide a diagram for this problem.

Here is the attached diagram.

SHARKYKESA wrote: Simple angle chasing. Firstly, we have $\angle EBC = \angle BEC = \angle BPC$, so $\angle PCB = 180^{\circ} - 2\angle EBC - \angle ABE$. Thus, we have $\angle PAB = 180^{\circ} - \angle PEB = 180^{\circ} \angle PCB = 2\angle EBC + \angle ABE$. We note that $\angle PBC = \angle DBF$ and $\angle EBC = \angle DBQ$. Therefore, $\angle DFQ = \angle DBQ = \angle EBC$, and $\angle DFB = \angle DBF = \angle EBC + \angle PBE$. Therefore, $\angle QAB = 180^{\circ} - \angle QFB = 180^{\circ} - (\angle DFQ + \angle QFB) = 180^{\circ} - (2 \angle EBC + \angle PBC) = 180^{\circ} - \angle PAB$. Therefore, $\angle PAQ = 180^{\circ}$, so $A$, $P$ and $Q$ are collinear. Is there any other method for proving this result?

John-Wick wrote: SHARKYKESA wrote: Simple angle chasing. Firstly, we have $\angle EBC = \angle BEC = \angle BPC$, so $\angle PCB = 180^{\circ} - 2\angle EBC - \angle ABE$. Thus, we have $\angle PAB = 180^{\circ} - \angle PEB = 180^{\circ} \angle PCB = 2\angle EBC + \angle ABE$. We note that $\angle PBC = \angle DBF$ and $\angle EBC = \angle DBQ$. Therefore, $\angle DFQ = \angle DBQ = \angle EBC$, and $\angle DFB = \angle DBF = \angle EBC + \angle PBE$. Therefore, $\angle QAB = 180^{\circ} - \angle QFB = 180^{\circ} - (\angle DFQ + \angle QFB) = 180^{\circ} - (2 \angle EBC + \angle PBC) = 180^{\circ} - \angle PAB$. Therefore, $\angle PAQ = 180^{\circ}$, so $A$, $P$ and $Q$ are collinear. Is there any other method for proving this result? Bashing

@above I guess bashing shouldn't be done for such an easy problem!!

Solution: Construction :Join $AE,AC,PE,PC,AF,BQ$ Here we can see that \[\angle CBE=\angle CEB \text{ (Since given that BC=CE)}\]we can see that \[\angle CBE=\angle CPE=\angle CAE=\alpha\text{ (Since we know that CBPE and CBAE is cyclic)}\]\[\angle CEB=\angle CAB=\alpha \text{ (since CBAE is cyclic)}\]now we can see in another circle that \[\angle FBD=\angle BFD=\beta \text{ (Since BD=DF)}\]and we can see here that line QE and line CD intersect at point B so we can say that \[\angle CBE=\angle QBF=\alpha\]\[\angle QBD=\angle QFD=\beta \text{ (Since QBFD is cyclic)}\]and here we can see that in $\triangle BPE$ have exterior angle $\angle PBQ$ so we can say that \[\angle BEP+\angle BPE=\angle PBQ\implies \angle BEP+2\alpha=\alpha+\beta\implies \angle BEP=\beta-\alpha\]so we can see here that $BEAP$ is cyclic quadrilateral so we can say that \[\angle BEP=\angle BAP=\beta-\alpha\]and we can see that \[\angle DFB-\angle DFQ=\beta-\alpha\]and since we know that $Q,B,A,F$ lie on circle $\omega_2$ so we can say that $\angle BAQ$ is $\beta-\alpha$ since we can see here that $\angle BAP=\angle BAQ$ and we are done !$\blacksquare$

does this work?

∠CEB = ∠CBE = ∠QBD = ∠QFD and ∠DFB = ∠DBF = ∠CEP ∠PAB = ∠PEB = ∠PEC - ∠BEC = ∠DFB - ∠DFQ = ∠QFB = ∠QAB so Q,P,A are collinear.

$\color{blue} \boxed {\textbf{SOLUTION}}$ $AP \cap \omega_2 \equiv Q'$ We want to show $Q \equiv Q'$ $\textbf{Step 1 : }$ $\angle CEP=\angle PBD=\angle FBD=\angle BFD$ $\angle CEB=\angle CBE=\angle QBD=\angle QFD$ $\textbf{Step 2 : }$ $\angle BFQ=\angle BFD - \angle QFD=\angle CEP - \angle CEB =\angle BEP=\angle BAP=\angle BAQ'=\angle BFQ'$ So, $FQ \equiv FQ'$ As, $Q$ and $Q'$ both lies on $\omega_2$ and $F$ is also on $\omega_2$, $\implies Q \equiv Q'$ $\blacksquare$

To avoid configuration issues let $\measuredangle$ be a directed angle modulo $\pi$. Also, let $\measuredangle DBF = \measuredangle BFD = \alpha$ and $\measuredangle EBC = \measuredangle CEB = \beta$. Note that \begin{align*} \measuredangle PCD = \measuredangle PCE - \measuredangle BCE = \measuredangle CEB + \measuredangle EBC - \measuredangle EBP = \beta - \alpha \\ \measuredangle QDC = \measuredangle DQB + \measuredangle QBD = \measuredangle EBC - \measuredangle BFD = \beta - \alpha \end{align*}So $CP \parallel QD$ and thus, by the converse of Reim's theorem (or simple angle chasing), we conclude that $Q - A - P$.

Angle chase

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); 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 = -15.28, xmax = 15.36, ymin = -8.88, ymax = 8.96; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); /* draw figures */ draw(circle((-6,-1), 5.168558793319468), linewidth(1) + blue); draw(circle((4,0), 7.65341638476964), linewidth(1) + ccqqqq); draw((-9.783183088799102,-4.521580002871505)--(10.205488237608218,-4.479586821523434), linewidth(1)); draw((-9.783183088799102,-4.521580002871505)--(-10.557747413852905,1.4374040517561189), linewidth(1)); draw((-10.557747413852905,1.4374040517561189)--(1.603952459053537,-7.268682035980443), linewidth(1)); draw((-2.96,3.18)--(-2.1934123141811837,-4.496268037518368), linewidth(1)); draw((-2.1934123141811837,-4.496268037518368)--(3.3249944877483792,7.623591667782995), linewidth(1)); draw((3.3249944877483792,7.623591667782995)--(10.205488237608218,-4.479586821523434), linewidth(1)); draw((-2.96,3.18)--(1.603952459053537,-7.268682035980443), linewidth(1) + linetype("4 4")); draw((3.3249944877483792,7.623591667782995)--(1.603952459053537,-7.268682035980443), linewidth(1)); draw((-10.557747413852905,1.4374040517561189)--(-0.8635976763904211,-1.5756484778214144), linewidth(1)); draw((-10.5,-1.64)--(-9.82,-1.64), linewidth(1)); draw((-10.5,-1.86)--(-9.82,-1.86), linewidth(1)); draw((-6.475575499931073,-4.28013728777642)--(-6.475575499931073,-5.000117339767044), linewidth(1)); draw((-6.210010795541072,-4.320554086134865)--(-6.210010795541072,-5.040357440247402), linewidth(1)); draw((6.615779589588047,0.48722713274971746)--(7.5001993819101065,1.5798386093215109), linewidth(1)); draw((4.757719405262763,-4.001111239961445)--(4.757719405262763,-5.258639480686026), linewidth(1)); /* dots and labels */ dot((-6,-1),dotstyle); label("$O_{1}$", (-5.92,-0.8), NE * labelscalefactor); dot((4,0),dotstyle); label("$O_{2}$", (4.08,0.2), NE * labelscalefactor); dot((-2.96,3.18),dotstyle); label("$A$", (-3.12,3.44), NE * labelscalefactor); dot((-2.1934123141811837,-4.496268037518368),dotstyle); label("$B$", (-2.3,-5.04), NE * labelscalefactor); dot((-9.783183088799102,-4.521580002871505),dotstyle); label("$C$", (-9.7,-4.32), NE * labelscalefactor); dot((10.205488237608218,-4.479586821523434),dotstyle); label("$D$", (10.28,-4.28), NE * labelscalefactor); dot((-10.557747413852905,1.4374040517561189),dotstyle); label("$E$", (-10.48,1.64), NE * labelscalefactor); dot((1.603952459053537,-7.268682035980443),dotstyle); label("$Q$", (1.68,-7.06), NE * labelscalefactor); dot((3.3249944877483792,7.623591667782995),dotstyle); label("$G$", (3.4,7.82), NE * labelscalefactor); dot((-0.8635976763904211,-1.5756484778214144),linewidth(4pt) + dotstyle); label("$P$", (-0.64,-1.76), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] USING ABOVE DIAGRAM. $\color{red}\textbf{Claim:-}$ $A,P,Q$ are collinear. $\color{blue}\textbf{Proof:-}$ First we observe that the points $E,A,P,B,C$ and $G,A,B,Q,D$ are cyclic points. Therefore we get some angle condition, $$\angle CBE=\angle CEB=\angle QBD=\angle QGD=2\angle EPB=2\angle EAB.$$Also we get, $$\angle EBP+\angle BEP=\angle ECB$$Now we get, $$\angle PEB=\angle PEC-\angle BEC\implies \angle BGQ=\angle BGD-\angle QGD$$$$\boxed{\angle QAB=\angle QGB.}$$Therefore the points $A,P,Q$ are collinear.