Dear friend from the beautiful Montevideo, one quick thought: ♦ All in this figure is symmetric, with symmetry axis the line (e) which is defined by the bisector of the angle $B$. $L$ is the polar of point $B$ with respect of the circle (g). Hence, this point is also the meeting point of the diagonals of the isosceles trapezium $TSQ_1Q_2$, with $S$ the midpoint of the arc $CSA$ and $T$ the midpoint of the arc $A_1TC_1$. We have here: $\angle CQ_1 S = \angle SQ_1 A,\;\;A_1 T = TC_1 \Rightarrow \angle TCB$ $ = \angle TAB = \frac{{\angle A + \angle C}} {4}\; \Rightarrow T \equiv P,\;\;Q_1 \equiv Q,$ because $\angle ATC = 90^ \circ + \frac{{\angle B}} {2}$

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WLOG $AB>BC$. By the hypothesis we get $\widehat{PAC}+\widehat{PCA}=\dfrac{1}{2}\cdot (\widehat{A}+\widehat{C})$, so $\widehat{APC}=90^\circ +\widehat{B}/2$. Therefore $AIPC$ is cyclic where $I$ is the incenter of $\triangle ABC$. We know that $[BL$ passes through the center $W$ of circle $(AIC)$. Hence, the reflection $M$ of $A$ about $BL$ lies on the circle $(AIC)$ and on the line $BC$. The reflection $N$ of $C$ about $BL$ also lies on the circle $(AIC)$ and on the line $AB$. Let $R$ be the reflection of $P$ about line $BL$. By the symmetry we get $\widehat{RMC}=\widehat{PAN}=\widehat{PCB}$. Therefore $PC\parallel RM$, so $arc\ RP=arc\ CM$ and it follows that $\overarc{AR}=\overarc{RC}$ since $AM\parallel CN\parallel PR$. Hence $R$ is the midpoint of arc $AC$ and $P$ is the midpoint of the arc $MN$. Let $I_b$ be the $B$-excenter of $\triangle ABC$. The division $(I,I_n,B,L)$ is harmonic, so $L$ lies on the polar of $B$ with respect to the circle $(AIC)$. Now take $R_1\in BR\cap (AIC),\ R_1\neq R$ and $P_1\in BP\cap (AIC),\ P_1\neq P$. Take $K\in RP_1\cap PR_1$. Obviously $K$ lies on $BL$ and it's well known that $K$ lies on the polar of $RR_1\cap PP_1$ with respect to the circle $(AIC)$. Hence $K=L$ and it follows that $R_1=Q$ since $Q\in PL\cap (AIC)$. Now, obviously $[QB$ is the bisector of $\widehat{AQC}$ since $B,Q,R$ are collinear.

Find a point $X$ in the plane such that $\angle ABX=\angle CBP$ (where the line $BX$ cuts segment $AC$) and $\angle AXB=\angle PCB$, and prove $X\equiv Q$.

o(center of circumcircle of triangle APL) is midpoint of AC in circumcircle of ABC. let angle bisector of B meet AC at T. OT.TB=AT.CT=PT.TQ ----> OPBQ is cycle now it is easy by play with angle.

Let $M$ be the midpoint of the arc $AC$.Now,it is easy to obtain $LA*LC=LB*LM=LP*LQ$,so we obtain $BPMQ$ is a cyclic and the rest is easy angle chasing.

Claim: Let $C'$ be the reflection of $C$ over the angle bisector of $\angle ABC$. Then, $BPCC'$ is cyclic. Proof.Indeed, \begin{align*} \measuredangle PCC'&=\measuredangle ACC'+\measuredangle PCA=\frac{\measuredangle B+\measuredangle C}{2}-\frac{\measuredangle B+\measuredangle C}{4}\\&=\frac{\measuredangle B+\measuredangle C}{4}=\measuredangle PBA=\measuredangle PBC'.\,\square \end{align*} Let $Q'$ be image of $P$ of inversion at $B$ with radius $\sqrt{ac}$ followed by the reflection over the angle bisector of $\angle ABC$. Thus, $Q'B$ is the angle bisector of $\angle CQ'A$ and $APCQ'$ is cyclic quadrilateral. Now, all we need is to show that $P,L,Q'$ are collinear. By ratio lemma,\begin{align*} \frac{AL}{LC}=\frac{AB}{BC}=\frac{AB}{BC}\cdot \frac{\sin{\angle ABP}}{\sin{\angle PBC}}\cdot \frac{\sin{\angle ABQ'}}{\sin{\angle Q'BC}}=\frac{AP}{PC}\cdot \frac{AQ'}{Q'C}, \end{align*}we are done. $\blacksquare$