[asy][asy] defaultpen(fontsize(9pt)); size(5cm); pair A,B,C,D,E,P,I,J,K,Q,X,M; B=dir(90); A=dir(210); C=dir(330); D=(3A+C)/4; E=(A+3C)/4; P=(9B+7D)/16; I=incenter(A,B,C); J=-P+2*foot(P,B,I); K=-J+2*foot(J,C,I); Q=extension(C,K,B,P); X=-Q+2*foot(Q,B,I); M=(A+C)/2; draw(B--A--C--cycle); draw(B--D); draw(B--E); draw(A--P); draw(C--Q); draw(A--Q); draw(A--X); draw(C--P); draw(C--X); draw(B--M,dotted); dot("$B$" , B , dir(B)); dot("$A$" , A , dir(A)); dot("$C$" , C , dir(C)); dot("$P$" , P , dir(45)); dot("$Q$" , Q , dir(120)); dot("$P^*$" , X , dir(30)); [/asy][/asy] Let $\ell^*$ be the line passing through $B$ such that $\{\ell,\ell^*\}$ are Isogonal Lines WRT $\Delta ABC$. Let $P^*$ be the reflection of $Q$ over the perpendicular bisector of $BC$. Due to symmetry note that $P^*$ lies on $\ell^*$. Some notation here, Directed angles are denoted by $\measuredangle$ and are assumed to be positive when taken in the clockwise direction. So, $\measuredangle CAP^*=\measuredangle QCA=\measuredangle PAB$. Hence, $P^*$ is the Isogonal Conjugate of $P$ WRT $\Delta ABC$. Now, $\measuredangle BCP=\measuredangle P^*CA=\measuredangle CAQ$. So, $$\measuredangle QAP=\measuredangle CAB-\measuredangle CAQ-\measuredangle PAB=\measuredangle BCA-\measuredangle BCP-\measuredangle QCA=\measuredangle PCQ. \blacksquare$$

Let $\ell$ cut circle $AQC$ at $R$ Then $\angle QRA = \angle PRA = \angle BAP$ so $BA$ is a tangent to $RAP$, by pop we have $BC$ is tangent to $PRC$ and now it's obvious that $\angle QAP = \angle PCQ$

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Claim: Let $Q'$ be the reflection of $Q$ over $B-$angle bisector, then $(P, Q')$ are isogonal conjugate pairs.} $\newline$ $\underline{\textbf{Proof:}}$ Note that $\angle BAP = \angle QCA = \angle Q'AC$ which means that lines $(AP, AQ')$ are isogonal, as well as $(BP, BQ')$ are isogonal. Hence, $(P, Q')$ are isogonal conjugates $\blacksquare$ $\newline$ Similarly $(Q, P')$ are also isogonal conjugates, where $P'$ is the reflection of $P$ over $B-$angle bisector. $\newline \rule{\textwidth}{0.5pt} \newline$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -5.697385686486056, xmax = -0.3783684148304967, ymin = 2.715611398192712, ymax = 5.895911542341867; /* image dimensions */ pen ttffqq = rgb(0.2,1,0); pen ffzztt = rgb(1,0.6,0.2); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-3.5,5.5)--(-5,3), linewidth(0.8) + blue); draw((-5,3)--(-2,3), linewidth(0.8) + blue); draw((-2,3)--(-3.5,5.5), linewidth(0.8) + blue); draw((-5,3)--(-3.5982914418216025,3.946754990784901), linewidth(0.8) + ttffqq); draw((-3.6106746569084267,3.7510699273125745)--(-2,3), linewidth(0.8) + ffzztt); draw((-3.4017085581783983,3.946754990784901)--(-2,3), linewidth(0.8) + ffzztt); draw((-5,3)--(-3.389325343091574,3.7510699273125745), linewidth(0.8) + ttffqq); draw((-3.5,5.5)--(-3.4017085581783983,3.946754990784901), linewidth(0.8)); draw((-3.4017085581783983,3.946754990784901)--(-5,3), linewidth(0.8) + ffzztt); draw((-2,3)--(-3.389325343091574,3.7510699273125745), linewidth(0.8) + ttffqq); draw((-5,3)--(-3.6106746569084267,3.7510699273125745), linewidth(0.8) + ffzztt); draw((-3.5982914418216025,3.946754990784901)--(-2,3), linewidth(0.8) + ttffqq); draw((-3.4017085581783983,3.946754990784901)--(-3.389325343091574,3.7510699273125745), linewidth(0.8) + wrwrwr); draw((-3.5982914418216025,3.946754990784901)--(-3.4017085581783983,3.946754990784901), linewidth(0.8)); draw((-3.6106746569084267,3.7510699273125745)--(-3.389325343091574,3.7510699273125745), linewidth(0.8)); draw((-3.5,5.5)--(-3.6582033762195567,3), linewidth(0.8)); /* dots and labels */ dot((-3.5,5.5),dotstyle); label("$B$", (-3.510061351536316,5.5452452600721465), NE * labelscalefactor); dot((-5,3),dotstyle); label("$A$", (-5.065491989722798,2.948231605242923), NE * labelscalefactor); dot((-2,3),dotstyle); label("$C$", (-1.96851848690507,2.9621193787981595), NE * labelscalefactor); dot((-3.6582033762195567,3),dotstyle); label("$D$", (-3.735737671808908,2.937815775076496), NE * labelscalefactor); dot((-3.5982914418216025,3.946754990784901),linewidth(4pt) + dotstyle); label("$P$", (-3.6906024077543895,3.968982961552805), NE * labelscalefactor); dot((-3.6106746569084267,3.7510699273125745),linewidth(4pt) + dotstyle); label("$Q$", (-3.7010182379208167,3.677339716892839), NE * labelscalefactor); dot((-3.4017085581783983,3.946754990784901),linewidth(4pt) + dotstyle); label("$P'$", (-3.364239729206333,3.98981462188566), NE * labelscalefactor); dot((-3.389325343091574,3.7510699273125745),linewidth(4pt) + dotstyle); label("$Q'$", (-3.360767785817524,3.694699433836884), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note that $\angle BAP' = \angle QAC$ or $$\angle BAP + \angle PAP' = \angle QAQ' +\angle Q'AC$$$$\boxed{\angle PAP' = \angle QAQ'}$$Similarly, $\angle PCP' = \angle QCQ'.$ But $\angle QAQ' = \angle QCQ'$ due to symmetry. Hence, $$\boxed{ \angle PAP'= \angle QAQ' = \angle QCQ' = \angle PCP' }$$Also that $\angle Q'AC =\angle BCP'$ because both the points are reflection over angle bisector and $\triangle BAC$ is isoceles. (It can also be visualized as mirror image with mirror kept at east i.e. near $C$) $\newline \rule{\textwidth}{0.5pt} \newline$ As $\triangle BAC$ is isoceles, $$\angle BAP + \angle PAQ + \angle QAQ' + \angle Q'AC = \angle BCP' + \angle P'CP + \angle PCQ + \angle QCA$$. Therefore, $$\boxed{\angle PAQ = \angle PCQ}$$

Since $AB=BC,$ it suffices to prove $\angle CAQ=\angle BCP.$ Let $Q'$ be the reflection of $Q$ with respect to $m,$ the perpendicular bisector of $\overline{AC}.$ Notice $Q'$ is the isogonal conjugate of $P$ with respect to $\triangle ABC$ as $m$ is the angle bisector of $\angle ABC$ and $AQ'QC$ being a cyclic isosceles trapezoid implies $\angle PAB=\angle QCA=\angle CAP'.$ Hence, $\angle CAQ=\angle P'CA=\angle BCP.$ $\square$

Considering: $(B,Q;P,L)=\frac{\sin(\angle PAB)}{\sin(\angle PAQ)} \div \frac{\sin (\angle LAB)}{\sin(\angle LAQ)} =\frac{\sin(\angle PCB)}{\sin(\angle PCQ)} \div \frac{\sin (\angle LCB)}{\sin(\angle LCQ)}$ Note that: $\angle LCB=\angle ACB=\angle BAC=\angle LAB$ and $\angle LCQ=\angle QCA=\angle BAP=\angle PAB$ $\implies \frac{\sin(\angle LAQ)}{\sin(\angle PAQ)}=\frac{\sin(\angle PCB)}{\sin(\angle PCQ)}$ But, $\angle LAQ+\angle PAQ=\angle PCB+\angle PCQ$ $\implies$ $\angle LAQ=\angle PCB$ and $\angle PAQ=\angle PCQ$.$\blacksquare$

Almost the same solution as above Let $R=\ell \cap AC$ $\frac{\sin{(\alpha)}\sin{(\beta_{1})}}{\sin{(\theta_{1})}\sin{(\alpha+\theta_{1}+\beta_{1})}}$ $=$ $\frac{(BD)(ER)}{(DE)(BR)}$ $=$ $\frac{\sin{(\beta_{2})}\sin{(\alpha)}}{\sin{(\theta_{2})}\sin{(\beta_{2}+\theta_{2}+\alpha)}}$ $\alpha+\theta_{1}+\beta_{1} = \beta_{2}+\theta_{2}+\alpha$ because the triangle is isoscoles ($AB=BC$) Then $\frac{(\beta_{1})}{(\theta_{1})}$ $=$ $\frac{(\beta_{2})}{(\theta_{2})}$ We have $\alpha+\theta_{1}+\beta_{1} = \beta_{2}+\theta_{2}+\alpha \rightarrow \theta_{1}+\beta_{1} = \beta_{2}+\theta_{2}$ Therefore $\beta_{1} = \beta_{2}$ and $\theta_{1} = \theta_{2}$

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