We dilate the figure with centre $A$ and ratio $\frac{1}{2}$. Let $M$ be midpoint of $BC$ and $J$ be midpoint of $AI$. Let $K$ be feet of $\perp$ from $A$ to $NJ$. Then $P\mapsto M$ and $R\mapsto K$. Also $Q\mapsto N$ and $AI\mapsto AI$. So we need to prove that $AI$ is tangent to $\odot(MNK)$. Claim 1 $AI$ is tangent to $\odot(NKI)$. Proof Consider $\triangle JAN$. $\angle JAN=\angle AKN=90^{\circ}\implies AJ^2=JK.JN$ $\implies JI^2=JK.JN$. So $JI$ is tangent to $\odot(NKI)$ which is same as our claim. Remark Infact $K$ is the $N-Humpty$ point of $\triangle NAI$. Claim 2 $AI$ is tangent to $\odot(MIN)$. Proof Let $N'$ be midpoint of arc $\overarc{BC}$ of $\odot(ABC)$ not containing $A$. Then $N'I^2=N'B^2=N'M.N'N$. So $N'I$ ie $AI$ is tangent to $\odot(MIN)$ as desired. Now $\exists$ unique circle passing through $N$ and $I$ tangent to $AI$. So $\odot(MIN)$ and $\odot(NKI)$ are one and the same. So $MNIK$ is cyclic and $\odot(MNIK)$ is tangent to $AI$. Hence $\odot(MNK)$ is tangent to $AI$ as desired.

Let $D$ be the midpoint of arc $\overarc{BC}$, $E$ the midpoint of $AR$ and $M$ the midpoint of $BC$. Consider the inversion $(D,DB)$, this maps $M\mapsto N$ and $I\mapsto I$, so $AI$ is tangent to $IBM$. Now $\angle DMI=\angle DIN=\angle IAN+\angle ANI=90^\circ +\angle ANI$, also $\angle IER=\angle ANI$ as $AQI\simeq RAI$ and $N,E$ are homologous, therefore $\angle IEN=\angle IER+\angle REN=\angle ANI+90^\circ =\angle DMI$, so $MIEN$ is cyclic. Now homotety $(A,2)$ finishes the problem.

Let $M$ be the midpoint of $BC$, let incircle touch $BC$ at $D$, let $F$ be point diametrically opposite to $D$ on incircle. Let $G=AF\cap BC$, let $O$ be the circumcentre of $\triangle ABC$ and let $E$ be the midpoint of arc $BC$ (not containing point $A$). Let $K=NE\cap AF$. Lemma. $\angle INA=\angle IMB$. Proof. For obvious reasons, $M$ is the point of $DG$ (comes from homothety centred at $A$, sending incircle to excircle). Now, we get that $\triangle KMG\sim \triangle FDG$, because $\angle KMG=\angle FDG=90^\circ$ (note that $NE$ is the diameter of circumcircle of $\triangle ABC$) and $\angle KGM=\angle FGD$. Since $GD=2GM$, we get that $FD=2KM$, but since $FI=ID$, then $KM=ID$, which implies that $IKMD$ is a rectangle. Thus, $IKNA$ is cyclic and because $IK\parallel BC$ and $KG \parallel IM$, hence $$\angle INA=\angle IKA=\angle AGB=\angle IMB. \quad \square$$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); 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 = -3.683422650571191, xmax = 0.8569444608186267, ymin = -1.689329481641562, ymax = 3.6378160461623725; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((-2.037124712344542,2.1799222783125747)--(-2.424776994598582,-0.3310072771965493)--(0.22711702718474006,-0.3045764397701375)--cycle, linewidth(1) + zzttqq); /* draw figures */ draw((-2.037124712344542,2.1799222783125747)--(-2.424776994598582,-0.3310072771965493), linewidth(1) + zzttqq); draw((-2.424776994598582,-0.3310072771965493)--(0.22711702718474006,-0.3045764397701375), linewidth(1) + zzttqq); draw((0.22711702718474006,-0.3045764397701375)--(-2.037124712344542,2.1799222783125747), linewidth(1) + zzttqq); draw(circle((-1.1094855736707452,0.7513190012203426), 1.703356009446683), linewidth(1)); draw((-1.0925094466317973,-0.9519524116874435)--(-2.037124712344542,2.1799222783125747), linewidth(1)); draw((-1.098829983706921,-0.31779185848334335)--(-1.5169551708003837,0.45529863437660134), linewidth(1)); draw((-1.1264617007096933,2.4545904141281287)--(-2.037124712344542,2.1799222783125747), linewidth(1)); draw(circle((-1.3217084357550386,1.454944524252365), 1.018534826449923), linewidth(1) + linetype("2 2")); draw(circle((-1.5169551708003837,0.45529863437660134), 0.7772192513081666), linewidth(1)); draw((-1.5169551708003837,0.45529863437660134)--(-1.509209184245518,-0.3218820166282464), linewidth(1)); draw((-1.5169551708003837,0.45529863437660134)--(-1.1065759702617868,0.4593887925215044), linewidth(1)); draw((-1.5247011573552491,1.2324792853814492)--(-1.5169551708003837,0.45529863437660134), linewidth(1)); draw((-0.6884507831683228,-0.31370170033844047)--(-2.037124712344542,2.1799222783125747), linewidth(1)); draw((-1.0925094466317973,-0.9519524116874435)--(-1.1264617007096933,2.4545904141281287), linewidth(1)); /* dots and labels */ dot((-2.037124712344542,2.1799222783125747),dotstyle); label("$A$", (-2.1508438602645192,2.3019318599592324), NE * labelscalefactor); dot((-2.424776994598582,-0.3310072771965493),dotstyle); label("$B$", (-2.3639296813766766,-0.28788042740391123), NE * labelscalefactor); dot((0.22711702718474006,-0.3045764397701375),dotstyle); label("$C$", (0.31603429953391965,-0.45179259749018613), NE * labelscalefactor); dot((-1.5169551708003837,0.45529863437660134),linewidth(4pt) + dotstyle); label("$I$", (-1.4869995714151054,0.5644628570447183), NE * labelscalefactor); dot((-1.1264617007096933,2.4545904141281287),linewidth(4pt) + dotstyle); label("$N$", (-1.0854147547037316,2.564191332097272), NE * labelscalefactor); dot((-1.098829983706921,-0.31779185848334335),linewidth(4pt) + dotstyle); label("$M$", (-1.0034586696605943,-0.5419442910376373), NE * labelscalefactor); dot((-1.0925094466317973,-0.9519524116874435),linewidth(4pt) + dotstyle); label("$E$", (-1.0526323206864767,-0.8861598482188147), NE * labelscalefactor); dot((-1.1065759702617868,0.4593887925215044),linewidth(4pt) + dotstyle); label("$K$", (-1.0034586696605943,0.3103989934109922), NE * labelscalefactor); dot((-1.1094855736707452,0.7513190012203426),linewidth(4pt) + dotstyle); label("$O$", (-1.0362411036778492,0.6792013761051108), NE * labelscalefactor); dot((-1.509209184245518,-0.3218820166282464),linewidth(4pt) + dotstyle); label("$D$", (-1.7246722180402043,-0.48457503150744113), NE * labelscalefactor); dot((-1.5247011573552491,1.2324792853814492),linewidth(4pt) + dotstyle); label("$F$", (-1.4951951799194192,1.3020676224329553), NE * labelscalefactor); dot((-0.6884507831683228,-0.31370170033844047),linewidth(4pt) + dotstyle); label("$G$", (-0.5854826359405929,-0.5255530740290099), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Main proof. Let $T$ be reflection of $A$ from $I$, which means that $IA=IT$. Because $\angle IRA=\angle IAQ=90^\circ$ and since $\angle AIR=\angle QIA$, thus $IAR\sim IAQ$, hence $IT^2=IA^2=IR\cdot IQ$. Now, if we manage to show that $T$ lies on $(PQR)$, then $AI$ is tangent to $(PQR)$. Since $I$ is the midpoint of $AT$, $M$ is the midpoint of $AP$ and $N$ is the midpoint of $AQ$, then $NM\parallel QP$, $IM\parallel TP$ and $IN \parallel QT$. Hence, $\angle IMO=\angle TPQ$. Now, $$90^\circ- \angle IMO=\angle IMB=\angle INA=90^\circ- \angle AIN=90^\circ- \angle ITQ\implies \angle IMO=\angle ITQ.$$Since, $IT^2=IR\cdot IQ$ and because $\angle IRT=\angle ITQ$, thus $\triangle IRT\sim\triangle ITQ$, hence $\angle IRT=\angle ITQ$. Furthermore, $\angle IRT=\angle TPQ\implies \angle TRQ+\angle TPQ=180^\circ$. $\quad \blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); 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 = -3.2872735182659847, xmax = 1.7854307157411335, ymin = -2.3809684782681813, ymax = 3.5707603161445003; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((-2.037124712344542,2.1799222783125747)--(-2.5661341718126205,0.2965707156783589)--(0.22711702718474006,-0.3045764397701375)--cycle, linewidth(1) + zzttqq); /* draw figures */ draw((-2.037124712344542,2.1799222783125747)--(-2.5661341718126205,0.2965707156783589), linewidth(1) + zzttqq); draw((-2.5661341718126205,0.2965707156783589)--(0.22711702718474006,-0.3045764397701375), linewidth(1) + zzttqq); draw((0.22711702718474006,-0.3045764397701375)--(-2.037124712344542,2.1799222783125747), linewidth(1) + zzttqq); draw(circle((-0.9819356361106197,0.8675613202024428), 1.6839582153293295), linewidth(1)); draw((-1.3362350603578086,-0.7787032967268194)--(-2.037124712344542,2.1799222783125747), linewidth(1)); draw((-1.1695085723139402,-0.004002862045889307)--(-1.712822951090764,0.810965721234017), linewidth(1)); draw((-0.6276362118634311,2.513825937131705)--(-2.037124712344542,2.1799222783125747), linewidth(1)); draw(circle((-1.712822951090764,0.810965721234017), 0.6824147775852571), linewidth(1)); draw((0.22711702718474006,-0.3045764397701375)--(-0.3018924322833385,-2.1879280024043535), linewidth(1)); draw((-0.3018924322833385,-2.1879280024043535)--(-2.5661341718126205,0.2965707156783589), linewidth(1)); draw((-0.3018924322833385,-2.1879280024043535)--(-1.1695085723139402,-0.004002862045889307), linewidth(1)); draw((-0.6276362118634311,2.513825937131705)--(0.7818522886176797,2.847729595950835), linewidth(1)); draw((-1.712822951090764,0.810965721234017)--(0.7818522886176797,2.847729595950835), linewidth(1)); draw((-1.2367734859561959,1.199633688002176)--(-2.037124712344542,2.1799222783125747), linewidth(1)); draw(circle((1.350557480786483,0.09088880354409788), 2.8148884081393994), linewidth(1)); draw((-1.388521189836986,-0.5579908358445407)--(-1.3362350603578086,-0.7787032967268194), linewidth(1)); draw((-1.1695085723139402,-0.004002862045889307)--(-2.037124712344542,2.1799222783125747), linewidth(1)); draw((-0.3018924322833385,-2.1879280024043535)--(-1.388521189836986,-0.5579908358445407), linewidth(1)); draw((-0.3018924322833385,-2.1879280024043535)--(0.7818522886176797,2.847729595950835), linewidth(1)); draw((-0.6276362118634311,2.513825937131705)--(-1.1695085723139402,-0.004002862045889307), linewidth(1)); /* dots and labels */ dot((-2.037124712344542,2.1799222783125747),dotstyle); label("$A$", (-2.1610233002463537,2.316319016399058), NE * labelscalefactor); dot((-2.5661341718126205,0.2965707156783589),dotstyle); label("$B$", (-2.4998140162359985,0.3385137554865363), NE * labelscalefactor); dot((0.22711702718474006,-0.3045764397701375),dotstyle); label("$C$", (0.3295462875694155,-0.46725875821856516), NE * labelscalefactor); dot((-1.712822951090764,0.810965721234017),linewidth(4pt) + dotstyle); label("$I$", (-1.675728490855781,0.9336866349278045), NE * labelscalefactor); dot((-0.6276362118634311,2.513825937131705),linewidth(4pt) + dotstyle); label("$N$", (-0.5861042961863819,2.636796720713587), NE * labelscalefactor); dot((-1.1695085723139402,-0.004002862045889307),linewidth(4pt) + dotstyle); label("$M$", (-1.0622425997393967,-0.24750261811717386), NE * labelscalefactor); dot((-0.9819356361106197,0.8675613202024428),linewidth(4pt) + dotstyle); label("$O$", (-0.8974254946633531,0.7871825415268769), NE * labelscalefactor); dot((-0.3018924322833385,-2.1879280024043535),dotstyle); label("$P$", (-0.26562659187185284,-2.097116797303884), NE * labelscalefactor); dot((0.7818522886176797,2.847729595950835),dotstyle); label("$Q$", (0.8148410969599882,2.938961413353), NE * labelscalefactor); dot((-1.2367734859561959,1.199633688002176),linewidth(4pt) + dotstyle); label("$R$", (-1.1995901873027663,1.2724773509174494), NE * labelscalefactor); dot((-1.388521189836986,-0.5579908358445407),dotstyle); label("$T$", (-1.300311751515904,-0.6412323691321666), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

One solution for this problems.

Attachments:

We do $\sqrt{bc}$ inversion and reflection and then scale from $A$ by a factor of $2$ from $A$. Now, the new question becomes, let $I_A$ be the $A-excenter$ and $A'$ be the reflection of $A$ over $I_A$, $P'$ is the point on $(ABC)$ such that $(AP';BC)=-1$. $Q'$ is the point at which exterior angle bisector of $\angle BAC$ intersects $BC$ and $R'$ is a point such that $AA'R'Q'$ is a rectangle and we want to show that the circle $(P'Q'R')$ is tangent to $AI_A$. We claim that the required point of tangency is $I_A$. So, as $AA'R'Q'$ is a rectangle and $I_A$ is midpoint of $AA'$, we have that $(R'Q'I_A)$ is tangent to $AI_A$. Now, let $M=AI_A\cap (ABC)$. Then, we know that $MP'Q"$ are collinear. Proof is because $(AP';BC)=-1\implies (AI_A\cap BC, MP'\cap BC; BC)=-1\implies MP'\cap BC=Q'$. Thus, $MP'\cdot MQ'=MB^2=MI_A^2$. So, $MI_A$ is tangent to $(I_AP'Q')$. Thus, $AI_A$ is tangent to $(P'Q'I_A)$ and $(Q'R'I_A)$. Thus, we are done.

Solution. Let $D$ be the projection of $I$ onto $BC$. We show that $S=\overline{AI}\cap \overline{PD}$ is the required tangency point. Firstly, we prove the following Claim 1. Define $M$ as the midpoint of $\overline{BC}$. Then $IM\parallel PD$. Proof. Recall that the symmetric of $D$ with respect to $I$ (say $D'$) lies on the incircle and the $A$-Nagel cevian of $\bigtriangleup ABC$, which also contains the symmetric of $D$ with respect to $M$ (say $D''$). Hence, $PD\parallel AD''\parallel IM$, since $D$ and $D''$ are corresponding points for $\bigtriangleup ABC$ and $\bigtriangleup PCB$. $\square $ Claim 1 implies that $AD'SD$ is a parallelogram, therefore $AI=IS$. Thus, $$IS^2=IA^2=IR\cdot IQ$$leading to conclude that $AI$ is tangent to $(QRS)$ at $S$ (1). Clearly, it also gives that $IN\parallel QS$ and we also have that $NM\parallel QP$. Then, by defining $T=\overline{NI}\cap (ABC),\ T\neq A$ and taking into account that $AT$ and $AD'$ are isogonal conjugates of $\bigtriangleup ABC$, it turns out that $$\angle SQP=\angle INM=\angle TNM = \angle TAI=\angle IAD'=\angle ISD$$which implies that $AI$ is tangent to $(SQP)$ at $S$. By (1) and the latter, we finally infer that $PQRS$ is a cyclic quadrilateral and that $AI$ is tangent to $(PQRS)$ at $S$. The result follows. $\blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12.411521472608563cm); 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 = -8.55461671777555, xmax = 32.81712152425299, ymin = -11.657425194610513, ymax = 9.079020623521094; /* image dimensions */ pen wqwqwq = rgb(0.3764705882352941,0.3764705882352941,0.3764705882352941); /* draw figures */ draw(circle((9.708747839497967,0.3484989496636267), 6.205224343661667), linewidth(1.) + wqwqwq); draw((7.06,5.96)--(3.88,-1.78), linewidth(1.) + wqwqwq); draw((3.88,-1.78)--(15.5,-1.88), linewidth(1.) + wqwqwq); draw((15.5,-1.88)--(7.06,5.96), linewidth(1.) + wqwqwq); draw((7.06,5.96)--(12.32,-9.62), linewidth(1.) + wqwqwq); draw((7.06,5.96)--(9.655348574654324,-5.856495625167647), linewidth(1.) + wqwqwq); draw((7.06,5.96)--(12.464294208683217,7.146987048989801), linewidth(1.) + wqwqwq); draw((9.762147104341608,6.5534935244949)--(9.655348574654324,-5.856495625167647), linewidth(1.) + wqwqwq); draw((8.138850453137717,1.0480464442392226)--(12.464294208683217,7.146987048989801), linewidth(1.) + wqwqwq); draw((7.06,5.96)--(10.095610589847283,3.807107600966758), linewidth(1.) + wqwqwq); /* dots and labels */ dot((7.06,5.96),linewidth(4.pt) + dotstyle); label("$A$", (7.149447737175381,6.1708605392709295), NE * labelscalefactor); dot((3.88,-1.78),linewidth(4.pt) + dotstyle); label("$B$", (3.9884041673382367,-1.5673741196903768), NE * labelscalefactor); dot((15.5,-1.88),linewidth(4.pt) + dotstyle); label("$C$", (15.59575615578023,-1.6685275139251652), NE * labelscalefactor); dot((8.138850453137717,1.0480464442392226),linewidth(4.pt) + dotstyle); label("$I$", (8.236846725199358,1.239632570324999), NE * labelscalefactor); dot((9.69,-1.83),linewidth(4.pt) + dotstyle); label("$M$", (9.779435987279886,-1.617950816807771), NE * labelscalefactor); dot((12.32,-9.62),linewidth(4.pt) + dotstyle); label("$P$", (12.409424237384389,-9.406762172886472), NE * labelscalefactor); dot((9.655348574654324,-5.856495625167647),linewidth(4.pt) + dotstyle); label("$L$", (9.754147638721188,-5.6640865861993035), NE * labelscalefactor); dot((9.217700906275436,-3.863907111521555),linewidth(4.pt) + dotstyle); label("$J$", (9.324245713223336,-3.6663070500622346), NE * labelscalefactor); dot((9.762147104341608,6.5534935244949),linewidth(4.pt) + dotstyle); label("$N$", (9.855301032955976,6.752492556120963), NE * labelscalefactor); dot((12.464294208683217,7.146987048989801),linewidth(4.pt) + dotstyle); label("$Q$", (12.561154328736572,7.359412921529692), NE * labelscalefactor); dot((10.095610589847283,3.807107600966758),linewidth(4.pt) + dotstyle); label("$R$", (10.209337912777736,4.021350911781678), SE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Construct $M$ midpoint of $BC$, $L$ midpoint of arc $\overarc{BC}$ and $J$ symmetric of $A$ over $I$. Since $R$ is the foot of $A$ in the right-angled triangle $\bigtriangleup AQI$, we have $IJ^2=IA^2=IR \cdot IQ$, which implies $AI$ is tangent to $\odot(QRJ)$ at $J$. By Shooting Lemma, we have $LM \cdot LN = LI^2$, so $\odot(NIM)$ is tangent to $AI$, at $I$, and after a homothety at $A$ with factor $2$ we get that $\odot(PJQ)$ is tangent to $AI$ at $J$. These two tangencies imply that $\odot(QRJ)$ and $\odot(PJQ)$ are the same circle which corresponds to $\odot(PQR)$, which is thus tangent to $AI$ at $J$. $\Box$

Let $O$ be the circumcenter of triangle $ABC$. Note that $N$ is the midpoint of $AQ$ and $AR \perp IQ$, so $N$ is circumcenter of triangle $ARQ$. Suppose that $AI$ intersects circumcircle of triangle $ABC$ at $T$, so points $T$, $O$ and $N$ are collinear, so $\angle QAI$ $=$ 90$^{\circ}$. Also, $AR \perp QI$ and by Euclides's Theorem in the triangle $AIQ$, $AI^{2}$ $=$ $IQ \cdot IR$ Let $K$ be the reflection of $A$ with respect to $I$ and let $M$ be the midpoint of $BC$. Since $ABPC$ is a parallelogram the diagonals intersect at their midpoint that is $M$. Also, $IK^{2}$ = $IR \cdot IQ$, therefore $IK$ is tangent to circumcircle of triangle $KRQ$. As $\frac{AM}{MP}$ = $\frac{AN}{NQ}$ = $\frac{AI}{IK}$ $=$ $1$ so $MN \parallel PQ$. So that, $MN \perp BC$ then $PQ \perp BC$. Analogously $IM \parallel KP$, so $\angle IMA$ $=$ $\angle KPA$ and $\angle AMN$ $=$ $\angle APQ$. Hence, $\angle IMN$ $=$ $\angle KPQ$. We get $\angle NIQ$ = $\angle IQK$ $=$ $\angle IKR$, as $\angle AIR$ is external angle in the triangle $IRK$, $\angle AIN$ $=$ $\angle IRK$. It is a known fact that $T$ is circumcenter of triangle $BIC$. So that, $TB$ $=$ $TI$ $=$ $TC$. Since, $TC \perp NC$ y $NT \perp BC$ we get $TI^{2}$ $=$ $TC^{2}$ $=$ $TM \cdot TN$. Thus $TI$ is tangent to circumcircle of triangle $IMN$ at $I$. Also, the triangles $TIN$ and $TMI$ are similar. Then, $\angle TIN$ $=$ $\angle TMI$, where $\angle IRK$ $=$ $\angle AIN$ $=$ $\angle IMN$ $=$ $\angle KPQ$, therefore $PQRK$ is a cyclic quadrilateral and $AI$ is tangent to the circumcircle of triangle $PQR$ at $K$, as needed.

Attachments:

Let $T$ be the reflection of the point $A$ through $I$. The key trick is to prove that $AI$ tangents to the circle formed by points $R,T,Q$; and after that it is tangent to the circle formed by points $T,P,Q$, which implies that these circles are the same; and yields the conclusion. Now both steps are easy. For the first, notice that $AI^2 = IR \cdot IQ$, and just notice that $AI = IT$, so the first step is done. For the next one, take an homothety cenntered at $A$ with scalinng factor $\frac12$, notice that the circle $TPQ$ goes to the circle $IMN$, which is clearly tangent to line $AI$ since $LI^2 = LM \cdot LN$, where $L$ is the midpoint of arc $BC$ and $M$ is the midpoint of side $BC$. So we're donee.

Let $A'$ be the reflection of $A$ over $I$.Let $M$ be the midpoint of $BC$ and $L$ be the midpoint of minor arc $BC>$.Then $P$ is the reflection of $A$ over $M$. CLAIM : $AA'$ is tangent to $(A'RQ)$ This is because $$IA'^2=IA^2=IQ.IR$$ CLAIM : $AA'$ is tangent to $A'PQ$ Consider the homothety at $A$ with ratio $\frac{1}{2}$.Then $P\to M, A' \to I ,Q \to N$.So , for this claim, it suffices to prove that $AI$ is tangent to $(IMN)$.But we have that $$LM.LN+=LB^2=LI^2$$as desired. Summing up , we have that $AA'$( or $AI$) is tangent to $(A'PQR)$.

Let $M$ midpoint of $BC$, $L$ midpoint of $\overarc{BC}$, $K$ reflection of $A$ over $I$. $$ LI^2=LB^2=LM\cdot LN $$Hence $(MIN)$ tangent to $AI$, and by homothety from $A$, $(PKQ)$ is tangent to $AI$. Also, $$ IK^2=IA^2=IR\cdot IQ, $$so $(KRQ)$ tangent to $AI$. Hence $(PRQ)$ tangent to $AI$ at $K$. $\blacksquare$

Let $E$ be the reflection of $A$ with respect to $I$ and $M=AP\cap BC$, the midpoint of $AP$ . Then we have that $IE^2=IA^2=IR \times IQ$, hence $IE$ is tangent to $(ERQ)$. Now, let $Z = AI \cap MN$, the midpoint of the small arc $BC$ in $(ABC)$, which is the center of $(BIC)$. Then we have that $\angle NMC = 90 = \angle NCZ => ZM \times ZN = ZC^2 = ZI^2$ and hence $ZI$ is tangent to $(MIN)$, i.e. $AI$ is tangent to $(MIN)$. Hence, by homothety, $AI$ is tangent to the circle passing through the reflections of $A$ with respect to $M,I,N$, which is $(PEQ)$. Hence, we conclude that $AI$ is tangent to both $(ERQ)$ and $(PEQ)$, so the two circles coincide. Hence, $P,E,R,Q$ are concyclic, in a circle tangent to $AI$, which proves that $AI$ is tangent to $PQR$, as needed.

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Gaussian_cyber wrote: Let $\triangle ABC$ be an acute-angled triangle with its incenter $I$. Suppose that $N$ is the midpoint of the arc $\overarc{BAC}$ of the circumcircle of triangle $\triangle ABC$, and $P$ is a point such that $ABPC$ is a parallelogram.Let $Q$ be the reflection of $A$ over $N$ and $R$ the projection of $A$ on $\overline{QI}$. Show that the line $\overline{AI}$ is tangent to the circumcircle of triangle $\triangle PQR$ Proposed by Patrik Bak - Slovakia Make o Homothety with center $A$ and radio $1/2$ Then we get that: $KI^2=KA^2=KR'*KN$ so $KI$ is tangent to $(R'NI)$ we are going to prove that $M$ belongs to $(R'IN)$ It is well know that $<ANI=<IMD$ so we have that: $<KR'I=<R'NI+<R'IN=<R'IK+<R'IN=90-<ANI=90-<IMD=<IMN$ done

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Let $M$ be the arc-midpoint of $\overarc{BC}$, $M'$ be the midpoint of ${BC}$ and $A'$ be the reflection of $A$ over $I$. Since $R$ is the foot of $A$ in the right-triangle $AQI$, we have $AI^2=IR \cdot IQ=A'I^2$, which implies $AI$ is tangent to $(A'QR)$ at $A'$. By EGMO Lemma 4.33$\big/$Shooting Lemma, we have $MI^2 \cdot MM' = MN^2$, so $(NIM')$ is tangent to $AI$, at $I$. The homothety at $A$ with ratio $\frac{1}{2}$ takes $Q \to N, P\to M', A' \to I$. This gives us that $AI$ is tangent to $(IM'N)\implies $ $AI$ tangent to $(QA'P)$ at $A'$. By both the tangency conditions we have that, $AI$ is tangent to $(A'PQR)$ at $A'$.

Similar to most of the above ig. Take the homothety centered at $A$ with factor $\frac{1}{2}$. Then if the midpoint of $BC$ is $M$ , midpoint of $AI$ is $D$ and foot of $A$ onto $DN$ is $E$, then it suffices to prove that $AI$ is tangent to $(NEM)$. We claim that the tangency point is in fact $I$. Firstly notice that since $\Delta ADN$ is right angled at $A$ and $E$ is foot of $A$ onto hypotenuse $$\implies DI^2=DA^2=DE.DN$$so $AI$ is tangent to $(NEI)$. Let the other midpoint of arc $BC$ be $M_A$. Since $\Delta M_ACN$ is right angled at $C$ and $M_AC=M_AI$ $$\implies M_AI^2=M_AC^2=M_AM.M_AN$$hence $M_AI$ which is equivalent to $AI$ is tangent to $(MNI)$. Now it is easy to see that if two circles with two common points have a common tangent through one of their intersection points, the circles must in fact be coincident. Hence $(NEIM)$ is cyclic and $AI$ is tangent to it at $I$.

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