Does EG stand for easy geometry? Let the bisector intersect the circumcircle at $M$ and $AB$ at $D$. And let Omega intersect the circumcircle at $T$ and $AB$ at $S$. From homothety follows that $M$ , $S$ and $T$ are colinear. By shooting lemma and PoP $MD\cdot MC=MS\cdot MT=MQ\cdot MP=MB^2$, thus $MB$ is tangent to the circumcircle of $BQP$, which implies that $$\angle{MBA}+\angle{AMQ}=\angle{MBQ}=\angle{BPM}=\angle{PBM}+\angle{PCB} \ \ \ \blacksquare$$

If we let $M$ denote the midpoint of arc $\widehat{AB}$ then the inversion at $M$ with radius $MA = MB$ fixes $\Omega$, so it swaps $P$ and $Q$, thus $\angle MPB = \angle QBM$. But $\angle MPB = \tfrac12 \angle C + \angle CBP$ and $\angle QBM = \angle QBA + \tfrac12 \angle C$, implying desired isogonality.

Note that isogonal conjugate of a point on the bisector is a projective map. Since vertex of the bisector swaps with the foot, incenter and $C$-excenter are fixed; we see that this projective map coincides with inversion $\odot(AIB)$. As $P, Q$ are inverses, we get $\angle ABP=\angle QBC$ as desired.

Let I be the incenter of $ABC$ and $M$ be the midpoint of arc $AB$ and let the tangency point of $\Omega$ and $AB,\Gamma$ be $K,L$ respectively,then obviously,$K,L,M$ are collinear.So by POP: $$MK.ML=MP.MQ=MB^2=MI^2$$Which implies that $I$ is the foot of angle bisector of $B$ in triangle $\triangle BPQ$.so we are done.

Let $\Omega$ touches $AB, \Gamma$ at $K, T$. Let $TC\cap\Omega = L$ and $M$ be the midpoint of arc $AB$ not containing $C$. By homothety, $T,K,M$ are colinear and $KL\parallel CM$. By Desargues' Involution Theorem on $KTKL$ and line $CM$, there exists involution swapping $(P,Q), (AB\cap CM, C)$ and $(M, {\infty}_{CM})$ Pencil from B gives involution swapping $(BA, BC), (BM_C, B\infty_{CM}), (BP,BQ)$ but the first two pairs are isogonal w.r.t. $\angle ABC$ so we are done. EDIT : I have attached the link to my own article on Desargues' Involution Theorem.

P1 was also geometry. Geoff rule??

Another inversion approach: Let $D$ be the tangency point of $\Omega$ with $AB$, $M$ be the midpoint of arc $\widehat{AB}$ and $T$ be the tangency point of $\Gamma$ and $\Omega$. Inverting trough $D$ with radius $DA.DB$ (directed lenghts), $\Gamma$ is fixed. Thus, $\Omega$ becomes the tangent line to $\Gamma$ through $M$, which is parallel to line $AB$. Note that such inversion swaps $A$ and $B$, $M$ and $T$. By the inversion, we have that $\angle{ABP} = \angle{DP'A}$ and $\angle{CBQ} = \angle{B} - \angle{DBQ} = \angle{B} - \angle{DQ'A}$. Through easy angle chasing, we can see that $\angle{DP'A} + \angle{DQ'A} = \angle{B} \iff \angle{MP'A} = \angle{MAQ'}$, which is equivalent to $MA$ be tangent to $(AP'Q')$. By Archimedes's Lemma, we have that $MA^2 = MD.MT$. Moreover, as $P, Q, M$ are collinear, we know that $D, T, P', Q'$ are concyclic. So $MD.MT = MP'.MQ' \Rightarrow MA^2 = MP'.MQ'$, as we wanted to show. $\blacksquare$

Let $M$ be the midpoint of arc $AB$ not containing $C$, and let $CM$ meet $AB$ at $N$; let the circle be tangent to $AB$ at $X$ and the circle at $Y$; by shooting $MP\cdot MQ = MX\cdot MY = MN\cdot MC = MB^2$; hence $(PQB)$ and $(BNC)$ are tangent at $B$; let $BP, BQ$ meet $(BNC)$ at $U,V$, then $UV\parallel NC$ and the result follows.

Let the $\Omega$ intersect the circumcircle at $T$ and $AB$ at $K$ and let the angle bisector of $\angle BCA$ intersect $\Gamma $ at point $M$. By homothety $M,K,T$ are collinear . Because $ATCM$ is cyclic we have $\angle {ATM}=\angle{ACM}$ ,but $\angle {MAB}=\angle{MCA}$.So we have $\angle {ATM}=\angle{MAB}$ , so $MA$ is tangent to $(KAT)$. $MA^2=MK*MT$ , but $MX*MY=MQ*MP$ and $MA^2=MB^2$ , so $MB^2=MQ*MP$.Then we have $MB$ tangent to $(PQB)$, so $\angle MBQ= \angle MPB $. $\angle ABQ$ = $\angle MBQ$ - $\angle MBA$ = $\angle MPB$ - $\angle MCB$=$180$ -$\angle BPC$ - $\angle PCB$=$\angle PBC$ . So $\angle ABQ$=$\angle PBC$ which is similar to $\angle ABP = \angle QBC$

We know $\Omega$ is a $\text{curvilinear incircle}$ of $\triangle ABC.$ Let $M=CP\cap \Gamma,$ $T=\Omega\cap \Gamma,$ and $K=\Omega\cap AB.$ Lemma: $T,K,M$ are collinear. Proof:Angle-chasing and homotety From this lemma we can get $MB^2=MK\cdot MT=MQ\cdot MP,$ then $MB$ is tangent to $(PQB),$ We know $\angle PBC=\angle MPB-\angle MCB=\angle MBQ-\angle MBA=\angle ABQ.$ As desired.

Let $M_C$ be the second intersection of $\Gamma$ and the angle bisector of $\angle{BCA}$. Invert about $M_C$ with radius $M_CA$. Then $\Gamma$ and line $AB$ get swapped, while the center of $\Omega$ and the center of its image must lie on a line through $M_C$. But there is only one circle centered on this line in this position that is tangent to $AB$ and internally tangent to $\Gamma$, so $\Omega$ is fixed under this inversion. So the power of $M_C$ with respect to $\Omega$ is $M_CA$. But it is also $M_CP\cdot M_CQ$, so $\triangle{M_CAP}\sim\triangle{M_CQA}$. So \begin{align*} \measuredangle{PAC}&=\measuredangle{BAC}-\measuredangle{BAP}\\ &=\measuredangle{BAC}-\measuredangle{BAM_C}-\measuredangle{M_CAP}\\ &=\measuredangle{BAC}-\measuredangle{BCM_C}+\measuredangle{APM_C}+\measuredangle{PM_CA}\\ &=\measuredangle{BAC}-\measuredangle{BCM_C}+\measuredangle{APM_C}+\measuredangle{CM_CA}\\ &=\measuredangle{BAC}-\measuredangle{BCM_C}+\measuredangle{APM_C}+\measuredangle{CBA}\\ &=\measuredangle{BCA}+\measuredangle{M_CCB}+\measuredangle{APM_C}\\ &=\measuredangle{M_CCA}+\measuredangle{APM_C}\\ &=\measuredangle{BCM_C}+\measuredangle{APM_C}\\ &=\measuredangle{BAM_C}+\measuredangle{M_CAQ}\\ &=\measuredangle{BAQ}, \end{align*}so $AP$ and $AQ$ are isogonal. But $CP$ and $CQ$ are isogonal, so $P$ and $Q$ are isogonal conjugates. Then $BP$ and $BQ$ are isogonal, so $\angle{ABP}=\angle{QBC}$.

Let $M_C$ denote midpoint of arc $AB$ not containing $C$. Its well-known that $\text{Pow}(M_C, \Omega)=M_CB^2$, so $M_CB^2=M_CP \cdot M_CQ$, thus by similarity $\angle M_CBP=\angle M_CQB$. But, $\angle M_CBP=\angle M_CBA+\angle ABP=\angle ABP+\frac{1}2 \angle C$ and $\angle QBM_C=\angle QBC+\angle QCB=\angle QBC+ \frac{1}2 \angle C$; thus $\angle ABP=\angle QBC$ as desired. $\square$

Denote by $T$ the tangent point between the two circles. By easy (trivial?) angle chasing and similar triangles it's enough to prove $MP \cdot MQ = MB^2$, where $M$ is the midpoint of arc $\widehat{AB}$. Now, do inversion around $M$ with radius $MA = MB$. After inversion, $A', B', C', T'$ lie on a line, and so do $M, C', P', Q'$. We also know that the circumcircle of $P'Q'T'$ is tangent to the line $A'B'$, and the same circle is tangent to the circumcircle of $A'B'M$. The condition to be proved is basically exactly the same as before: $MQ' \cdot MP' = MB'^2$. This is a well-known basic configuration: we have a circle $\Omega$ (circumcircle of $A'B'M$), a chord of $\Omega$ ($A'B'$), and a circle $\omega$ tangent both to the chord and $\Omega$ (circumcircle of $P'Q'T'$). The midpoint of the arc between the ends of the chord ($M$) is on the same line as the tangent points of $\omega$ with $\Omega$ and $A'B'$. From here it's power of a point and easy similar triangles to prove $MQ' \cdot MP' = MB'^2$.

Let $D$ be a point on segment $AB$ such that $CD$ is tangent to $\Omega$ at $Y$. Let $X$ be the tangency point on $AB$ and $C'=CI \cap AB$ and $M$ be the midpoint of arc $AB$ not containing $C$. It is well-known $XIY$ are colinear. Also note $\measuredangle CYI=\measuredangle IXC'$ as they both subtend arc $XY$ hence using sine rule and angle bisector theorem on $\triangle AC'C$ we see: $$\frac{C'X}{CY}=\frac{C'I}{CI}=\frac{AC'}{AC}$$Now by power of a point: $$\frac{C'P}{PC} \cdot \frac{C'Q}{QC}=\left (\frac{C'X}{CY} \right)^2=\left (\frac{AC'}{AC} \right)^2$$But it's well-known this means $P,Q$ are isogonal in $\triangle ACC'$ and hence by symmetry $P,Q$ are isogonal conjugates in $\triangle ABC$ as desired.

Let $X$ be the midpoint of the arc $AB$ not containing $C$. Also let $Y$ and $Z$ be the tangency point of $\Omega$ with $AB$ and $\Gamma$ respectively. $D$ is the feet of angle bisector of $\angle ACB$. $I$ denotes the incenter of $\triangle ABC$ Lemma 1: $X,Y,Z$ collinear. Proof: Let $O_1$ and $O_2$ denotes the circumcenters of $\Omega$ and $\Gamma$ respectively. We know that, $Z, O_1, O_2$ collinear. Also, $YO_1 \perp AB$ and $XO_2 \perp AB$ implies $YO_1 || XO_2$. Now, homothety gives us $X,Y,Z$ collinear. $\square$ Lemma 2: $Y,Z,C,D$ cyclic. Proof: $\angle YZC = \angle XZC = \angle XAC = \angle A + \frac {\angle C}{2}$ $\angle YDX = \angle A + \frac {\angle C}{2}$ $\square$ Lemma 3: $XB$ is tangent to $\bigodot PQB$ Proof: $\angle XBD = \angle DCB = \frac {\angle C}{2} \Rightarrow XB$ is tangent to $\bigodot BDC \Rightarrow XD.XC=XB^2$ $XP.XQ = XY.XZ = XD.XC = XB^2$ $\square$ Lemma 4: $BI$ bisects $\angle QBP$ Proof: Lemma 3 gives us $$XP.XQ=XB^2=XI^2$$$$\Rightarrow \frac{XP}{XI}= \frac {XI}{XQ}$$$$\Rightarrow \frac{XP-XI}{XI}= \frac {XI-XQ}{XQ}$$$$\Rightarrow \frac{PI}{XI}= \frac {QI}{XQ}$$$$\Rightarrow \frac{PI}{QI}= \frac {XI}{XQ} = \frac{XP}{XI} = \frac{XP}{XB}$$$$\triangle XPB \sim \triangle BPQ \Rightarrow \frac{XP}{XB}=\frac{BP}{BQ}$$$$\therefore \frac{PI}{QI} = \frac{BP}{BQ}$$$\square$ Now our problem statement follows immediately. $Q.E.D$

.Let $N$ be the intersection point of $(ABC)$ and the angle bisector of $\angle{ACB}$ other than $C$ Suppose $\Omega$ touches $\tau$ at $L$ and $AB$ at $M$ Let $M'=NL \cap AB$ and $XY$ be the common tangent of $\tau$ and $\Omega$ Define $K=XY \cap AB$ $KM=KL$ $\measuredangle KLM' = \measuredangle KLN = \measuredangle KLB + \measuredangle BLN = \measuredangle LAB + \measuredangle BAN = \measuredangle LNB + \measuredangle NBA = \measuredangle M'NB + \measuredangle NBM' = \measuredangle NM'B = \measuredangle LM'B = \measuredangle LM'K$ So, we get $KM' = KL$ which means $MM'=0$ By similarity we get $NB^2=NM.NL=NQ.NP$ So, $\triangle NBQ \sim -\triangle NPB$ $\measuredangle NBQ = \measuredangle BPN$ $\measuredangle NBA+ \measuredangle ABQ = \measuredangle BCN + \measuredangle PBC$ $\measuredangle ABQ = \measuredangle PBC$ $\measuredangle ABP = \measuredangle QBC$

Let $D$ be the second intersection of $\Gamma$ and the bisector $\angle BCA$.Let $E,F$ be the tangency point of $\Omega$ with $AB$ and $\Gamma$ respectively. $Lemma$ 1: $DE \cdot DF=DB^2$ $Proof$ : $$\angle DFB= \frac{\angle C}{2}= \angle DBE$$. So, $\triangle DFB$ is similar to $\triangle DBE$ $$\rightarrow DE \cdot DF=DB^2$$ $Lemma$ 2: $DP \cdot DQ=DB^2$ $Proof$ : By, POP we get, $$P_\Omega(D)=DP \cdot DQ=DE \cdot DF=DB^2$$ $Lemma$ 3: $BI$ is the angle bisector of $\angle PBQ$ where $I$ is the incentre of $\triangle ABC$ $Proof$ : From $Lemma$ $3$ we get $$DP \cdot DQ=DB^2$$And, $P,Q,D$ collinear. So, $D$ is the centre of the Appolonian Circle of $\triangle BPQ$. Let, the Appolonian Circle $(D,DB)$ intersect $PQ$ at $I$. So, $BI$ is the angle bisector of $\angle PBQ$. And from $Fact$ $5$ we get $I$ is the incentre of $\triangle ABC$. Now, $$\angle PBI=\angle QBI$$$$\rightarrow \angle PBI + \angle ABI=\angle QBI + \angle CBI$$$$\rightarrow \angle ABP= \angle QBC$$.

If $M$ is the midpoint of arc $AB$ not containing $C$, then $P, Q$ are swapped under an inversion at $M$ with radius $MA$. Thus $$\angle ABP = \angle MBP - \angle MBA = \angle MQB - \angle BCQ = \angle CBQ$$as desired.

let TANGENCY point of two circles $K$ and TANGENCY point of circle to $AB$ also $S$obviously $KS$and $PQ$ concurrent on midpoint of arc $AB$ then compute power of point M . and according this equality $\angle SBM=$ $\angle SKB$we get $MB^2=MS.MK=MP.MQ $ then with angle chasing we get the solution.

Let $M$ be the intersection of the angle bisector with circle $\Gamma$, $K$ be the tangency point of circle $\Omega$ with $AB$ and $T$ be the tangency point of circle $\Omega$ with circle $\Gamma$. By homothety $M-K-T$ are collinear. $$\angle MAB = \angle MTB = \angle MBA$$By converse alternate segment theorem $MB$ is tangent to $(KTB) \implies MB^2 = MK \cdot MT$ and again by POP we have $$MP \cdot MQ = MK \cdot MT \implies MB^2 = MP \cdot MQ$$hence $MB$ is tangent to $(BPQ)$ and from here the result follows.

Ah nice First note that $\overline{C,P,Q,M}$ and $\overline{X,D,M}$ where $M$ is the midpoint of arc $AB$. We get, $MD \times MX= MB^2 = MQ \times MP$ so $MB$ is tangent to $\odot{BQP}$ which implies the result on angle chase.

Let $M$ be midpoint of $\overarc{AB}$ and let $\omega$ be tangent to $\Gamma$ at $K$ $\text{EGMO Lemma 4.43}$ implies $M-D-K$ are collinear and $M$ obviously lies on the $C$ angle bisector

Consider the inversion with radius $MA=MB$ Observe that $$(AB \iff \odot(ABC)) \implies (\omega \iff \omega) \implies (P \iff Q) \implies MP \ \cdot \ MQ = MB^2 \implies \angle MQB = \angle MBP$$$$\angle MQB = \angle MBP \iff \angle MBA + \angle ABP = \angle QBC + \angle QCB = \angle QBC + \angle QCA \iff \angle ABP=\angle QBC \ \blacksquare$$

microsoft_office_word wrote: Let $\Gamma $ be the circumcircle of triangle $ABC$. A circle $\Omega$ is tangent to the line segment $AB$ and is tangent to $\Gamma$ at a point lying on the same side of the line $AB$ as $C$. The angle bisector of $\angle BCA$ intersects $\Omega$ at two different points $P$ and $Q$. Prove that $\angle ABP = \angle QBC$. Inversion makes me feel guilty some times,

BRUH , a P5? Let $M$ be the mid point of $\widehat{AB}$ .Let the tangency point with $AB$ and $\Gamma$ be $K$ and $T$, respc. Then $M-K-T$ and $C-P-Q-M$ .Also, $MP.MQ=MK.MT=MB^2=MA^2 \implies$ $MB$ is tangent to the circumcircle of $BPQ$.Thus $\angle MPB=\angle MBQ \implies \angle CBP=\angle ABQ$ $\blacksquare$

Fix $ABC$ and vary $P$ on bisector of angle $C$. Let $I,$ $J,$ $Q'$ be incenter, $C-\text{excenter}$ and isogonal conjugate of $P$ in $\triangle ABC.$ Clearly $P\mapsto Q'$ is an involution fixing $I,$ $J,$ so it's nothing but inversion wrt $\odot (AIBJ),$ and hence $Q=Q'$ as desired.

Let $M$ is the midpoint of minor arc $AC$. $BQ\cap \Gamma$, $MN\cap AB=X$. $\Omega$ is tangent to $\Gamma$ at $D$ and $AB$ at $E$. By shooting lemma, $MN.MX=MD.ME=MQ.MP\implies$ points $X,P,Q,N$ are concyclic. So, $\angle MBX =\angle MCA=\angle BCM =\angle BMN=\angle MPX\implies$ points $X,P,B,M$ are concyclic. So, $\angle QBC=\angle XMP=\angle XBP=\angle ABP$.

Suppose $\Omega$ touches $AB$ and $\Gamma$ at $K, T$ respectively, and let the $C$-bisector meet $\Gamma$ again at $M$. Because $\Omega$ is a Curvilinear Incircle wrt $\Gamma$, we know $M, K, T$ are collinear. Now, the Shooting Lemma implies $$MB^2 = MK \cdot MT = Pow_{\Omega}(M) = MP \cdot MQ$$so $MB$ is tangent to $(BPQ)$. This yields $$\angle ABP = \angle MBP - \angle MBA = \angle BQP - \angle MAB = \angle BQP - \angle MCB = \angle QBC$$as desired. $\blacksquare$ Remark: Let $X$ lie on $BC$ such that $KX \parallel PQ$. Then, Steiner's Ratio Lemma implies $CX$ is tangent to $(PQX)$.

Let $\Omega$ be tangent to $ABC$ and $AB$ at $R$ and $S$ and Let $T$ be midpoint of arc $AB$ not containing $C$. It's well Known that $R,S,T$ are collinear. $TP.TQ = TS.TR = TB^2 \implies \angle TQB = \angle PBT$ so $\angle CBQ = \angle TQB - \angle TCB = \angle PBT - \angle ABT = \angle PBA$.

Let $\Omega$ touch $\overline{AB}$ at $K$ and $\Gamma$ at $T$, and let $M$ be the arc midpoint of $\widehat{AB}$ not containing $C$. It is then well-known that $M,K,T$ are collinear and $MK \cdot MT=MB^2$, hence by PoP we find $MB^2=MP\cdot MQ$, i.e. $\overline{MB}$ is tangent to $(BPQ)$. Then, WLOG let $C,Q,P,M$ lie on $\overline{CM}$ in that order, so \begin{align*} \angle ABP&=\angle QBC &\iff\\ \angle MBP-\angle MBA&=\angle MBQ-\angle MBC &\iff\\ \angle MQB-\frac{\angle C}{2}&=\angle MBQ-\left(\angle B+\frac{\angle C}{2}\right) &\iff\\ \angle B+\angle C&=\angle MQB+\angle MBQ &\iff\\ \angle A&=\angle QMB=\angle CMB, \end{align*}which is evident. $\blacksquare$

Solved with mxlcv and jelena_ivanchic! Let $M$ be the midpoint of arc $AB$ not containing $C$. The angle bisector of $C$ passes through $M$. It's well known that $MA^2=MB^2$ is power of $M$ wrt $\Omega$. Hence $MA^2=MP\cdot MQ \implies MA$ tangent $(APQ) \implies \angle BAP=\angle MAP -\angle MAB=\angle AQP-\angle MCB=\angle QAC$. As $\angle BAP=\angle CAQ$ and $\angle ACP=\angle BCQ$, hence $P, Q$ are isogonal conjugate wrt $\triangle ABC$. Hence $\angle ABP = \angle QBC$, and we are done!

Let $T$ be the tangency point, and $M$ the arc midpoint of minor arc $\widehat{AB}$. It is well-known that $M, E, T$ are collinear and also $$MP \cdot MQ = ME \cdot MT = MB^2.$$As a result, $\triangle BQM \sim \triangle PBM$. Now angle chase: $$\measuredangle ABQ = \measuredangle ABM + \measuredangle MBQ = \measuredangle BPM + \measuredangle MCB = \measuredangle PBC,$$which suffices.

Easy. Solved on the plane . Let $M$ be the midpoint of $\overarc{AB}$. $L$ be intersection with $\Omega$ $F$ be circles' intersection . So we have $M,L,F$ collinear from shooting lemma. We have $MB^2=ML.MF=MP.MQ$ from PoP. So from similarities if we chasing angle we get desired result.

I literally forgot that there existed something called Shooting Lemma. :yaw: [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 */ /* A few re-additions (otherwise throws error due to the missing xmin) are done using bubu-asy.py. This adds back the dps, real xmin coordinates, changes the linewidth, the fontsize and adds the directions to the labellings. */ pair A = (-37.57730,75.56340); pair B = (-100.84251,-43.18470); pair C = (39.83866,-44.14172); pair D = (-83.79684,-11.19018); pair M = (-101.25884,33.26400); pair T = (-56.51806,-80.63562); pair P = (-60.13324,10.70260); pair Q = (7.98004,-26.66416); import graph; size(10cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); draw(arc(B,23.68739,-0.38976,8.63225)--B--cycle, linewidth(0.75) + blue); draw(arc(B,23.68739,52.93070,61.95272)--B--cycle, linewidth(0.75) + blue); draw(A--B, linewidth(0.5)); draw(B--C, linewidth(0.5)); draw(C--A, linewidth(0.5)); draw(circle((-30.23601,-4.57470), 80.47365), linewidth(0.5)); draw(circle((-40.49370,-34.26071), 49.06538), linewidth(0.5)); draw(M--T, linewidth(0.5) + blue); draw(M--C, linewidth(0.5)); draw(B--M, linewidth(0.5) + blue); draw(B--P, linewidth(0.5) + red); draw(B--Q, linewidth(0.5) + red); dot("$A$", A, N); dot("$B$", B, W); dot("$C$", C, dir(0)); dot("$D$", D, W); dot("$M$", M, NW); dot("$T$", T, SW); dot("$P$", P, N); dot("$Q$", Q, NE); [/asy][/asy] Let $T$ denote the tangency point of $\Omega$ and $\Gamma$ and $D$ denote the tangency point of $\Omega$ and $AB$. Now from Shooting Lemma, we have that $\overline{M-D-T}$ are collinear and that $MD\cdot MT=MB^2$. This further gives that $MB^2=MD\cdot MT=MP\cdot MQ$ which means that $\measuredangle MBP=\measuredangle BQP$. Now to finish, we have, \begin{align*} \measuredangle ABP&=\measuredangle MBP-\measuredangle MBA\\ &=\measuredangle BQP-\measuredangle MCA\\ &=\measuredangle BQP-\measuredangle BCM\\ &=\measuredangle BQC+\measuredangle QCB\\ &=\measuredangle QBC .\end{align*} This gives us what we wanted and we are done.

Let $S$ be the midpoint of $\overarc{AB}$,$T$ be the tangency point of both circles and $K$ be the touch point of $AB$ to $\Omega$.It's a well known lemma that $T,K,S$ are collinear.Let $\angle SCB=\alpha$ and $\angle SBP=\beta \implies \angle PBC=\beta- \alpha$. So we need $\angle ABQ=\beta-\alpha$. $\angle SBA=\angle BTK$ $\implies$ $BS^2=SK\cdot ST$ and $SK\cdot ST=SQ\cdot SP$. This finishes the problem.

Shooting lemma does it $MB^2=MP\cdot MQ$. Therefore we have , $\triangle BQM \sim \triangle PBM$. Now by angle chasing we are done

This took an embarassingly long time. Relabel $A, C$ so that they switch places. Suppose $\Omega$ is tangent to $BC$ and $\Gamma$ at $D$ and $T$ respectively. Let the angle bisector of $\angle BAC$ intersect $\Gamma$ at $M$. Note that $\angle MTC = \angle CBM = \angle BCM$ which implies $MC^2=MD \cdot MT$. Since $TDQP$ is cyclic, this implies $MP \cdot MQ = MD \cdot MT = MC^2$. Therefore, $\angle CAM + \angle ACP = \angle MPC = \angle MCQ = \angle BCQ + \angle BCM$ $\implies$ $\angle ACP = \angle BCQ$, as desired. $\blacksquare$

Let $\omega $ tangents $ \gamma$ at $T$ and $BC$ at $D$ Let $M=TD \cap \gamma $ From 5 stars lemma $MB$ tangents to $(BQP)$ and we are done

Lol the flagship shooting/archimedes lemma problem. Let the circles be tangent at $X$, let the circle be tangent to $BC$ at $D$ and let the midpoint of arc $AB$ not containing $C$ be $M$. Also I take $P$ to be nearer to $M$ than $Q$.(The same words can probably be copy pasted for both the cases but just in case.) Then by archimedes' lemma, $\overline{X-D-M}$ is collinear and $MD.MX=MP.MQ=MA^2=MB^2$. This implies that $MB$ is tangent to $(PBQ)$ so $\angle MBP= \angle MQB$. Note that $\angle MBP= \frac{C}{2} + \angle ABP$ and $\angle MQB= \frac{C}{2}+ \angle QCB$ so the desired result follows.