Suppose that $P$ is closer to $B$. Let the bisector of $\angle{PAQ}$ intersect $\overline{BC}$ at $H$. Let $\overline{PH}$ intersect $\overline{AQ}$ at $G$. Clearly $\overline{PG}$ is tangent to $\omega$. Note that the segment $QG$ is the reflection of $PR$ over $\overline{AH}$. $$\angle{PGA}=\angle{PGQ}=\angle{ARQ}=180-\angle{RPB}-\angle{RBP}=180-\angle{ACB}-(180-\angle{PBA}-\angle{ABC})=\angle{PBA}$$Thus $G$ lies on $\omega$, and the result follows from Poncelet's Porism.

Take AQ$\cap \omega$=K Angles : AKP = 180-ABP = APB+PAB=ABC+BAP= $\frac{arcAC+arcBP} {2}$= ARC So, the incircle of ARQ is also of AKP Since BC is tangent to $\omega$, tangents from $B$ and $C$ must lie on the circumcirlce (by Poncelet's porism)

My solution: Problem is equivalent to below:(We just reverse the problem statement) In $\triangle ABC$, $M$ is the midpoint of arc $BAC$.tangents from $M$ to incircle of $\triangle ABC$ intersect $BC$ and $(ABC)$ at $K,L,T,S$($K,L,M$ and $M,T,S$ are collinear,$K,T$ are on $BC$,$L,S$ are on $(ABC)$) then $ML=MT$ and $MK=MS$. Proof: We use following two lemmas: Lemma 1:$KLTS$ is cyclic. proof is trivial by inversion with center $M$ radius $MB^2$. Lemma 2:$LS$ is tangent to the incircle of $\triangle ABC$ This is well-known and true for any point on $(ABC)$ replaced with $M$. Back to OP,Let $LS \cap KT =X$,then by lemma 2,$MLXT$ is inscribed.Also,by lemma 1,$\angle MLX =\angle MTX$ which means that $MLXT$ is kite,finishing our proof.

govind7701 wrote: Take AQ$\cap \omega$=K Angles : AKP = 180-ABP = APB+PAB=ABC+BAP= $\frac{arcAC+arcBP} {2}$= ARC So, the incircle of ARQ is also of AKP By Poncelet's porism, we are done Can you elaborate where do you use Poncelet Porism?

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(30cm); 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 = -25.452355633764675, xmax = 7.206350371537186, ymin = -11.60879940294072, ymax = 9.842227310515382; /* image dimensions */ pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-14.074411844006573,6.1856950212950075)--(-13.624244305053052,6.248953392507186)--(-13.68750267626523,6.6991209314607065)--(-14.137670215218751,6.635862560248529)--cycle, linewidth(1.6)); 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dot((-11.666002891786134,4.517805346546533),linewidth(4pt) + dotstyle); label("$S$", (-11.587406299492825,4.699124002593838), NE * labelscalefactor); dot((-15.463483961649452,4.978940434021899),linewidth(4pt) + dotstyle); label("$K$", (-15.38044498908497,5.149145542036973), NE * labelscalefactor); dot((-12.406552331736776,5.408506093653417),linewidth(4pt) + dotstyle); label("$L$", (-12.316012601448378,5.577737484363769), NE * labelscalefactor); dot((-14.137670215218751,6.635862560248529),linewidth(4pt) + dotstyle); label("$J$", (-14.051809967871902,6.7992245199951356), NE * labelscalefactor); dot((-13.935018146693112,5.193723263837656),linewidth(4pt) + dotstyle); label("$N$", (-13.858943593824844,5.363441513200371), NE * labelscalefactor); label("$90^\circ$", (-13.901802788057523,6.306343786319321), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We begin with a crucial lemma Lemma $CJ \perp JG$ at $J$. Proof Let $LK \cap HS=N$ . By la-hires since $N$ lies on the polars of $J,G$ we have that $JG$ is precicely the polar of $N$. Now let $J'=GN \cap JG$ we have that $GJ'$ is tangent to $\odot(JKLG) \implies J\equiv J'$ and we are done with the claim.$\blacksquare$ Now back to the main problem. We use phantom points . Let $F=GJ \cap \omega$. We'll prove that $F \equiv J$. It is trivial to see that $GF$ bisects $\angle KFL$ . Now $\angle AFG=\angle CBG=90- \frac{\angle BGA}{2}=90-\frac{\angle BFA}{2}$ and so $\angle CFG=90^\circ \implies F \equiv J$ and we are done.$\blacksquare$.

Iran TST 2018 Day 2 P5 wrote: Let $\omega$ be the circumcircle of isosceles triangle $ABC$ ($AB=AC$). Points $P$ and $Q$ lie on $\omega$ and $BC$ respectively such that $AP=AQ$ .$AP$ and $BC$ intersect at $R$. Prove that the tangents from $B$ and $C$ to the incircle of $\triangle AQR$ (different from $BC$) are concurrent on $\omega$. Proposed by Ali Zamani, Hooman Fattahi Solution:- Notations:- Let $AQ\cap\odot(ABC)=T$ and let the incircle of $\triangle AQR$ be $\gamma$. Claim 1:- $PRTQ$ is a cyclic quadrilateral, moreover $PRTQ$ is an isosceles trapezoid. Let $\angle ABC=\theta$ and $\angle BAT=\alpha\implies\angle AQR=\theta+\alpha$. Also $\angle QAC=180^\circ-2\theta-\alpha=\angle CBT\implies\angle APT=\theta+\alpha$ as $AB=AC$.So $TQPR$ is a cyclic quadrilateral. So, $$AQ.AT=AP.AR\implies AT=AR\implies QT=PR$$Hence, $PRTQ$ is also an isosceles trapezoid. Claim 2:- $PT$ is tangent to $\gamma$. So, if $PT\cap QR=K$. Then $PK=QK$, hence by the converse of Pilot's Theorem $AQKP$ must have an inscribed circle which is $\gamma$. Hence, $PT$ is tangent to $\gamma$. So $\triangle APT$ has both an inconic as well as circumconic.So, by Poncelet's Porism we get that tangents at $B$ and $C$ to $\gamma$ must intersect on $\omega$. $\blacksquare$.

Let $I$ and $I_A$ be the incenter and $A-excenter$ of $\triangle ARQ$. Now notice that $\angle ARI=\angle IRQ=\angle II_AQ=\angle PI_AI$. Hence $I, I_A,P,R$ are concyclic. Now by shooting lemma, $$AB^2=AC^2=AR\times AP=AI\times AI_A$$Therefore $\angle ACI=\angle IBC$. $$\angle BIC=180^{\circ}-\angle ACB=\angle ACQ=90^{\circ}+\frac{A}{2}$$Suppose the tangents from $B$ and $C$ to the incircle meet at $X$. Then $I$ is the incenter of $\triangle XBC$. Hence $$\angle BXC=2(\angle BIC-90^{\circ})=\angle BAC$$Hence $B,A,X,C$ are concyclic.

Let $AQ$ intersect $\omega$ at $S$. Note that by the shooting lemma, $AQ \cdot AS = AB^2 = AP \cdot AR$, so $AS = AR$. Thus $PS$ and $QR$ are reflections across the angle bisector of $\angle{QAR}$, and thus $PS$ is also tangent to the incircle of $AQR$. This finishes the problem by poncelet's porism.

Consider the inversion $\varphi: \mathbb{R}^2\to \mathbb{R}^2$ centered at $A$ with power $AB^2$. Then $P$ and $R$ are interchanged and so are $Q$ and $\varphi(Q)=AQ\cap (ABC)\neq A$. Since $AP=AQ$, then $AR=AQ'$, and by symmetry $\triangle AQR$ and $\triangle AP\varphi(Q)$ share an incircle $\omega$. By Poncelet's Porism, there is a triangle with side $BC$ inscribed in $(ABC)$ and circumscribed about $\omega$. Hence done. $\blacksquare$