I'm not as smart as #2 so here's a stupid way to do it (which is basically the same).

3300th post, I guess

Invert with center $A$, radius $AI$. Clearly $B \rightarrow P$, $C \rightarrow Q$. Let the incircle be tangent to $AB, AC$ at $D, E$, then $AI^2 = AD \cdot AD' = AE \cdot AE'$ gives $\angle AID' = \angle AIE' = 90^{\circ}$. So the incircle maps into a circle tangent to $AB, AC$ at $D', E'$ which is the $A$-mixtilinear incircle, which is tangent to $(ABC)$, so the incircle is tangent to its inverse which is $PQ$.

This is true for any $P$ on $AB$ and $Q$ on $AC$ which satisfy $\angle{PIQ}=90^{\circ} - \tfrac{A}{2}.$ Consider $T,$ the intersection of the circumcircle of $\triangle{APQ}$ with $AI.$ Clearly, $T$ lies on the perpendicular bisector of $PQ,$ and $\angle{PTQ}=2\angle{PIQ},$ hence $T$ is the circumcenter of $\triangle{PIQ},$ implying that $I$ is the $A$-excenter of $\triangle{APQ}.$

http://www.artofproblemsolving.com/community/c4h1133016 lol

Additional property of the configuration: Prove that the lines $PQ$, $BC$ and the line of the centres of the circles $(BIP)$ and $(CIQ)$ are concurrent.

Quote: Let $ABC$ be a triangle with incentre $I$. The circle through $B$ tangent to $AI$ at $I$ meets side $AB$ again at $P$. The circle through $C$ tangent to $AI$ at $I$ meets side $AC$ again at $Q$. Prove that $PQ$ is tangent to the incircle of $ABC.$ Solution: $AP \cdot AB=AQ \cdot AC=AI^2$ $\implies$ $BPQC$ cyclic. Let $T$ be foot from $I$ to $PQ$ and Let $\Delta DEF$ be the intouch triangle in $\Delta ABC$ $\implies$ $TPFI$ cyclic $\implies$ $\angle TIF$ $=$ $\angle ACB$ but some simple angle chasing shows $\angle PIF$ $=$ $\frac{1}{2}$ $\angle ACB$ $\implies$ $PI$ bisects $\angle TIF$ $\implies$ $\Delta TPI$ $\cong$ $\Delta FPI$ $\implies$ $T$ $\in$ $\odot (DEF)$

silouan wrote: Additional property of the configuration: Prove that the lines $PQ$, $BC$ and the line of the centres of the circles $(BIP)$ and $(CIQ)$ are concurrent. Let $U$ be the intersection of the line through the aforementioned centers and $AB$. It is easy to deduce that $U$,$B$,$C$ and the center of $(QIC)$ are cocyclic as well as that $U, P, Q$ and the center of $(QIC)$ are cocyclic. Hence, since $BCPQ$ is cyclic we are done by radical axis theorem.

Let $E$ and $F$ be the tangency points of the incircle on $AC$ and $AB$. By angle chasing, \[ \angle PIF = \angle AIF - \angle AIP = \left( 90^{\circ} - \frac{1}{2} \angle A \right) - \frac{1}{2} \angle B = \frac{1}{2} \angle C. \]Similarly, $\angle EIQ = \frac{1}{2} \angle B$. [asy][asy] pair A = dir(110); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair D = foot(I, B, C); pair E = foot(I, C, A); pair F = foot(I, A, B); pair _P = abs(A-I)*abs(A-I)/abs(A-B) * dir(B-A) + A; pair _Q = abs(A-I)*abs(A-I)/abs(A-C) * dir(C-A) + A; pair P = _P; pair Q = _Q; filldraw(incircle(A, B, C), invisible, blue); filldraw(A--B--C--cycle, invisible, blue); draw(P--Q, red); draw(P--I--F, deepgreen); draw(E--I--Q, deepgreen); pair T = foot(I, P, Q); draw(I--T, red+dashed); draw(A--I, lightblue); draw(B--I--C, lightblue); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(270)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$T$", T, dir(T)); /* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ A = dir 110 B = dir 210 C = dir 330 I 270 = incenter A B C D = foot I B C E = foot I C A F = foot I A B !pair _P = abs(A-I)*abs(A-I)/abs(A-B) * dir(B-A) + A; !pair _Q = abs(A-I)*abs(A-I)/abs(A-C) * dir(C-A) + A; P = _P Q = _Q incircle A B C / 0.1 blue / blue A--B--C--cycle / 0.1 blue / blue P--Q / red P--I--F / deepgreen E--I--Q / deepgreen T = foot I P Q I--T / red dashed A--I / lightblue B--I--C / lightblue */ [/asy][/asy] Let $T_p$, $T_q$ denote the second tangency of $P$, $Q$ to the incircle. Then $\angle E I T_q = \angle B$ and $\angle T_p I F = \angle C$. Since $\angle E I F = \angle B + \angle C$, it follows $T_p = T_q$. Remark: Notice we really only need $\angle PIQ = 90^{\circ} - \frac{1}{2} \angle A$ by essentially the same argument.

Here's my solution: By simple angle chase, one sees that $\angle PIQ=90^{\circ}-\frac{A}{2}$. We prove that the problem is true for all points $P,Q$ satisfying $\measuredangle PIQ=90^{\circ}-\frac{A}{2}$, such that $P$ and $Q$ lie on the same side of the incircle $\omega$. Animate $P$ on $AB$. Then, by rotation centered at $I$, we see that $P \mapsto Q$ is projective. Also $P \mapsto Q'$ is also a projective map, where $Q'$ is the point where the tangent from $P$ to $\omega$ (other than $AB$) meets $AC$ (Just consider two pole-polar transformations). So it suffices to show that $Q \equiv Q'$ for three positions of $P$. But, the result is obvious when $P=A,B$ and $P=\omega \cap AB$. Hence, done. $\blacksquare$

It is enough to show that $PI$ bisects $B\hat{P}Q$ as then $I$ is equidistant from $PQ$ and $BP$ so in particular the perpendicular distance from $I$ to $PQ$ is the inradius $r$. By alternate segment theorem, $B\hat{P}I = 180 - A\hat{I}B = \tfrac{1}{2}(\hat{A} + \hat{B})$. $AP \cdot AB = AI^{2} = AQ \cdot AC$ so $PQCB$ is cyclic so $A\hat{P}Q = \hat{C}$ and hence $I\hat{P}Q = 180 - \hat{C} - \tfrac{1}{2}(\hat{A} + \hat{B}) = \tfrac{1}{2}(\hat{A} + \hat{B}) = B\hat{P}I$.

Finally after dumbly staring at it! We will show that $I$ is the $A$ excenter of $APQ$.By Pop $AI^2=AP.AB=AQ.AC \implies B,C,P,Q$ conyclic.Since $AI$ is the angle bisector of $\angle PAQ$ and Angle chasing gives $\angle PIQ=90^{\circ}-\angle \dfrac{A}{2}$ and $\angle BQI=90^{\circ}-\angle \dfrac{B}{2}$ so we conclude that $I$ is the excenter and everything follows easily that $PQ$ is tangent to the incircle. REMARKS-In my first attempt I was trying to invert around the incircle with power inradius.We get a following equivalent problem Inverted wrote: Let $DEF$ be the contact triangle of $ABC$.Let $\Gamma$ be the incircle.Let $P,Q,R$ be the midpoints of $EF,DF,DE$ repectively.Let $\alpha,\beta,\gamma$ be $\odot{IQR},\odot{IRP},\odot{IPQ}$.Let the line through $Q,R$ parallel to $AI$ cut $(IPQ),(IPR)$ at $U,V$ then $(IUV)$ is tangent to $\Gamma$.Similar variants also follow

Nice extension that could have been given in contest at #7

Nothing new.{almost similar #5.} we proof the general problem when $\angle PIQ= 90- \angle \frac{A}{2}$. Let the $(APQ)$ intersects $AI$ at $X$. Then $\angle XAP=\angle IAP=\angle IAQ=\angle XAQ \implies XP=XQ$ and $\angle PIQ=90-\angle \frac{A}{2}=\frac{1}{2}(180-\angle A)=1/2 (\angle PXQ) \implies X$ is the circumcenter of $\triangle PIQ$ and since $AI$ is the angle bisector by fact5/ Incenter excenter lemma I is the $A$ excenter of PAQ.

Let $AI$ meet arc $\widehat{BC}$ at $M$. Let $PQ$ meet $(ABC)$ again at $X,Y$. Inversion in $A$ radius $AI$ fixes $X,Y$, so Fact 5 implies that $I$ is the incenter of $\triangle MXY$. We're done by Poncelet's porism.

Inversion about $I$ also works alright, you just get you want to prove a circle has the same radius as the nine point circle of the intouch triangle, which is easy with a little angle chasing.

Using the fact that ABC is similar to triang APQ we get substitutions PB+QC=AB+AC-AI^2(AB+AC)/AB*AC. And we get that PQ+BC=BC*AI^2/AB*AC + BC. We use Pitot theorem in quadrilateral PQCB.In these given substitutions proceed by using ravi substitution for the sides of triangle ABC and we get Heron's Formula, which holds.Implying that a circle can be inscribed in quadrilateral PQCB and since the incircle touches PB,BC,QC we get that PQ is tangent to incircle.

Here's a computational approach (similar to #20). Let \(a=BC\), etc. Let \(s=\frac{a+b+c}{2}\). First, we compute the length of \(AI^2\). We can see by stewart's theorem that \(AD^2=\frac{bc((b+c)^2-a^2)}{(b+c)^2}\), and note that \(\frac{AI}{AD}=\frac{b+c}{a+b+c}\) by mass points, so we can compute \(AI^2=\frac{bc((b+c)^2-a^2)}{(a+b+c)^2}=\frac{bc(b+c-a)}{a+b+c}\). It suffices to show that the incenter of \(ABC\) is the excenter of \(APQ\). This is equivalent to showing that the tangency points of the \(A\)-excircle of \(APQ\) are the tangency points of the incenter of \(ABC\) on \(AB,AC\). Thus, it suffices to show that \(2(s-a)=\frac{AP}{b}(a+b+c)\) (because \(AQP\) and \(ABC\) are similar). Finally, by power of a point, we can show that \(AP=\frac{AI^2}{c}\), so the right hand side becomes \(\frac{AI^2(a+b+c)}{bc}\) whereas the left hand side is \(b+c-a\). The conclusion follows.

Let the perpendiculars from $I$ to $CA, AB$ be $E, F$, respectively; assume that $B, F, P, A$ and $C, E, Q, A$ lie in this order (the other configurations follow similarly). Let $X$ be the foot of the perpendicular from $I$ to $PQ$; it suffices to prove $X$ lies on the incircle, or that $\angle{FXE} = \frac{360-\angle{FIE}}{2} =\frac{180+A}{2}$. From cyclic quads $IXQE$ and $IFPX$ and the given tangencies we have \begin{align*} \angle{IXE} &= \angle{IQE} \\ &= 90-(\angle{AIE}-\angle{AIQ}) \\ &= 90-\left(90-\frac{A}{2}-\frac{C}{2}\right) \\ &= \frac{A}{2}+\frac{C}{2}\end{align*}and similarly $$\angle{IXF} = \frac{A}{2}+\frac{B}{2},$$so $$\angle{FXE} = \angle{IXE}+\angle{IXF} = \frac{180+A}{2},$$as desired.

found this in elem geo unit?? Invert about $A$ with radius $AI$ note that $PQ$ and incircle are sent to circumcircle and $A$-mixtilinear incircle, done! time to find elem geo sol. GRIND.

rmtf1111 wrote: Let $ABC$ be a triangle with incentre $I$. The circle through $B$ tangent to $AI$ at $I$ meets side $AB$ again at $P$. The circle through $C$ tangent to $AI$ at $I$ meets side $AC$ again at $Q$. Prove that $PQ$ is tangent to the incircle of $ABC.$

Let the intouch triangle be $\Delta DEF$. It suffices to prove that the reflection of $E$ across $IQ$ is equal to the reflection of $F$ across $IP$. Since we already have $IE=IF$, we just need $\angle EIF = 2 \angle PIQ$. This is clearly true because $\angle EIF = 180^\circ - \angle A = \angle B + \angle C$ and \[ \angle PIQ = \angle AIP + \angle AIQ = \angle ABI + \angle ACI = \frac{\angle B}2 + \frac{\angle C}2. \]

$AP\cdot AB=AI^2=AQ\cdot AC$ Let $B' $ and $C'$ be reflections of $B$and $C $ across angle bisector WE know that $B'C'$ is tangent to the excircle Therefore by homothety $PQ $ Maps to $B'C'$ and the incircle maps to the excircle. We are done

We know that \[\angle AIB = 90+\frac{\angle C}{2} \implies \angle BPI = \angle 180 - \angle AIB = 90-\frac{\angle C}{2}\]due to the tangency of $(BPI)$. Note it suffices to show that $PM$ is tangent to $(I)$ (as $QM$ tangency follows similarly), which is equivlent to showing: \[\angle BPI = \angle QPI \iff \angle BPC = 180 - \angle C\]However this follows since $AI,$ $BP,$ and $QC$ concur, radical axis theorem implies $BPQC$ cyclic, so we're done.

We have $AP \cdot AB = AI^2 = AQ \cdot AC$, so $\triangle APQ \sim \triangle ABC$. The unique point $X$ on the angle bisector of $\angle PAQ$ for which $\angle PXA$ = $\tfrac{\angle PQA}{2}$ is the $A$-excenter of $\triangle APQ$. Since $I$ satisfies this property, $X$ must be the $A$-excenter of $\triangle APQ$, hence $\overline{PQ}$ is tangent to the incircle of $\triangle ABC$.

Here's a solution with Inversion at $A$. Let the circle through $B$ tangent to $AI$ at $I$ be $\omega_B$ and the circle through $C$ tangent to $AI$ at $I$ be $\omega_C$. Invert at $A$ with radius $AI$. It is well known that this sends the incircle to the $A-$mixtilinear incircle. On the other hand, it preserves the circles $\omega_B$ and $\omega_C$, thus it must send $P \mapsto B$ and $Q \mapsto C$. Thus $\overleftrightarrow{PQ} \mapsto (ABC)$ which is tangent to the $A-$mixtilinear incircle. Thus, we are done. $\square$

Here's a solution with Inversion at $I$. Then the problem reduces to the following: (reduction is left as an exercise) Inverted Problem: Let $ABC$ be a triangle, let $AD$ be the $A-$altitude, $O$ be its circumcenter, $\omega$ be the circumcircle. Let $\ell_B$ be the perpendicular bisector of $BD$ and $\ell_C$ be the perpendicular bisector of $CD$. Suppose $M,N$ are the midpoints of $AB,AC$ respectively. Let $\ell_B$ intersect the circle with diameter $OB$ at $M,P$, and let $\ell_C$ intersect the circle with diameter $OC$ at $N,Q$. Prove that the circumcircle of $OPQ$ is tangent to $\omega$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.65586257191475cm); 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 = -0.6547086119207891, xmax = 26.00115395999396, ymin = -6.7892150984667134, ymax = 5.155792059246044; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); /* draw figures */ draw((6.64,3.8)--(3.36,-3.24), linewidth(0.8) + wrwrwr); draw(circle((7.79842031029619,-1.0238094627516343), 4.960955084279531), linewidth(0.8) + sexdts); draw((6.64,3.8)--(12.12,-3.46), linewidth(0.8) + wrwrwr); draw((12.12,-3.46)--(3.36,-3.24), linewidth(0.8) + wrwrwr); draw((3.36,-3.24)--(6.3993212811294615,-5.783388987754199), linewidth(0.8) + linetype("4 4") + wrwrwr); draw((6.3993212811294615,-5.783388987754199)--(12.12,-3.46), linewidth(0.8) + linetype("4 4") + wrwrwr); draw(circle((5.579210155148096,-2.131904731375819), 2.4804775421397656), linewidth(0.8) + wvvxds); draw(circle((9.959210155148094,-2.241904731375814), 2.480477542139765), linewidth(0.8) + wvvxds); draw((5.,0.28)--(4.8796606405647305,-4.5116944938770995), linewidth(0.8) + wrwrwr); draw((9.38,0.17)--(9.259660640564729,-4.6216944938771025), linewidth(0.8) + wrwrwr); draw((6.64,3.8)--(6.3993212811294615,-5.783388987754199), linewidth(0.8) + wrwrwr); /* dots and labels */ dot((6.64,3.8),linewidth(3.pt) + dotstyle); label("$A$", (6.7093623709817285,3.9111603438259017), NE * labelscalefactor); dot((3.36,-3.24),linewidth(3.pt) + dotstyle); label("$B$", (2.8026017086907213,-3.6430627067102415), NE * labelscalefactor); dot((12.12,-3.46),linewidth(3.pt) + dotstyle); label("$C$", (12.189199229150972,-3.349191329458263), NE * labelscalefactor); dot((6.461240330268538,-3.3178850311254653),linewidth(3.pt) + dotstyle); label("$D$", (6.536496854951153,-3.210898916633803), NE * labelscalefactor); dot((5.,0.28),linewidth(3.pt) + dotstyle); label("$M$", (4.669549281820937,0.4884231264205096), NE * labelscalefactor); dot((9.38,0.17),linewidth(3.pt) + dotstyle); label("$N$", (9.440637524264822,0.2809845071838192), NE * labelscalefactor); dot((6.3993212811294615,-5.783388987754199),linewidth(3.pt) + dotstyle); label("$R$", (6.467350648538923,-5.6828757958710305), NE * labelscalefactor); dot((4.8796606405647305,-4.5116944938770995),linewidth(3.pt) + dotstyle); label("$P$", (5.067139968691261,-4.3690978740386575), NE * labelscalefactor); dot((9.259660640564729,-4.6216944938771025),linewidth(3.pt) + dotstyle); label("$Q$", (9.336918214646476,-4.524676838466176), NE * labelscalefactor); dot((7.79842031029619,-1.0238094627516343),linewidth(3.pt) + dotstyle); label("$O$", (7.6946958123560085,-0.7216354857935179), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Proof: Let $AD$ intersect $\omega$ at $A, R$. Then the homothety at $B$ of ratio $2$ sends the circle with diameter $OB$ to $\omega$, therefore $P \mapsto R$. Thus, $P$ is the midpoint of $BR$. Similarly $Q$ is the midpoint of $CR$. Thus, the homothety at $R$ with ratio $2$ sends the circumcircle of $OPQ$ to $\omega$, hence they are indeed tangent at $R$. $\square$

Using power of point, $AB \cdot AP = AI^2 = AC \cdot AQ \implies BCPQ$ concyclic. $\measuredangle AQP = \measuredangle CBA = 2 \measuredangle IBA = 2 \measuredangle AIP$. Similarly, $\measuredangle APQ = \measuredangle AIQ$. Combining these with the fact that $AI$ is the angle bisector of $\angle PAQ$, we have that $I$ is the $A$-excentre of $\Delta APQ$. Since the incircle is tangent to lines $APB$ and $AQC$, it is also tangent to $PQ$.

Denote the circle passing through $B,I,P$ as $\omega_1$ and the circle passing through $C,I,Q$ as $\omega_2$. By POP on $\omega_1$, we get $AP \cdot AB = AI^2$, and by POP on $\omega_2$, we get $AQ \cdot AC = AI^2$, giving $AP \cdot AB = AQ \cdot AC$, which means $PQCB$ is a cyclic quad. Let the incircle of $\triangle ABC$ be tangent to $AB$ at $F$. Construct $IR$ perpendicular to $PQ$ at $R$. If we prove $IF = IR$, we will be done, as it implies that $IR$ is the radius of the circle and perpendicular to $PQ$, so $PQ$ is tangent to the incircle at $R$. Now, since $PQBC$ is cyclic, $\angle RPF = 180 - \angle C$, and since $\angle PFI = \angle PRI = 90$, we have $\angle RPF + \angle RIF = 180$, so $\angle RIF = \angle C$. Notice that since $AI$ is tangent to $\omega_1$, $\angle AIP = \angle ABI = \angle B/2$. $\angle AIB = 90 + \angle C/2$, so $\angle BIP = \angle AIB - \angle AIP = 90 + \angle C/2 - \angle B/2$. $\angle BIF = 180 - (90 - \angle B/2) = 90 - \angle B/2$. $\angle PIF = \angle PIB - \angle FIB = \angle C/2$, so $PI$ bisects $\angle RIF$. This means $\triangle PRI \cong \triangle PFI$, giving $IF = IR$ and we are done.

Trying to write this up on my phone oops $AP \cdot AB =AI^2= AQ \cdot AC$ implies $ABQP$ is a cyclic quadrilateral. Now $$\angle IQC = \angle QAI + \angle QIA= \tfrac{\angle A}{2} + \angle ICA = \tfrac{180^{\circ}- \angle B}{2}$$Therefore $QI$ bisects $\angle PQC$ so we can reach the desired tangency.

Wow, 20SLG8 configuration again! This was pretty easy especially if you have seen that problem before. Let $D$ and $E$ be the second intersections of circles $(BIP)$ and $(CIQ)$ with line $\overline{BC}$ respectively. Then, we start off with the following observation. Claim : Lines $\overline{DP}$ and $\overline{EQ}$ are both parallel to line $\overline{AI}$. Proof : This is quite clear. Note that, \[\measuredangle AIQ = \measuredangle ICQ = \measuredangle ECI = \measuredangle EQI\]from which it follows that $DP \parallel AI$. Similarly, we can also show that $EQ \parallel AI$, proving the claim. Now, note that \[AP \cdot AB = AI^2 = AQ \cdot AC\]so it follows that quadrilateral $BPQC$ is cyclic. This further implies that, \[2\measuredangle CQI = 2\measuredangle CQE = 2\measuredangle EQI = 2\measuredangle CAI + 2\measuredangle ECI = \measuredangle CAB + \measuredangle BCA = \measuredangle CBA = \measuredangle AQP\]which implies that $\measuredangle CQI = \measuredangle IQP$. Thus, $\overline{QI}$ is the internal $\angle CQP$-bisector. Similarly, we can show that $\overline{PI}$ bisects $\angle BPQ$. Then, in quadrilateral $BPQC$, all internal angle bisectors concur, at $I$. Thus, it must be a tangential quadrilateral with incenter $I$. Further, since the incircle of $\triangle ABC$ is centered at $I$ and is tangent to $BC$, the incircle of $BPQC$ must also precisely coincide with the incircle of $\triangle ABC$. This means $PQ$ is also tangent to the incircle of $\triangle ABC$, as we wished to show.

Too easy for $EGMO$, no? Note that the given condition can be restated to prove that the incircle of $\Delta ABC$ is the $A$ excircle of $\Delta APQ$. By some easy angle chasing, we get that $\angle PIQ$ is $90- \frac{A}{2}$ so if the $A$ excenter is $I'$, then $I$ lies on $(PI'Q)$ as well as $AI$ and it lies on the opposite side of $PQ$ as $A$. Note that these conditions uniquely determine the point so $I \equiv I'$. Now since the incircle is already tangent to the rays $AP, AQ$ ;we have the three parameters necessary and sufficient to determine the circle, so the incircle of $\Delta ABC$ is indeed the $A$ excircle of $\Delta APQ$.

Since $AI$ is the radical axis of $(PIB)$ and $(QIC)$ and $AI$, $PB$, and $QC$ all intersect at $A$, we must have $BPQC$ cyclic, giving $\triangle{AQP}\sim\triangle{ABC}$. Now since $AI$ is tangent to $(PIB)$, we have \[\angle{IPB}=180-\angle{AIB} = \frac{A+B}{2}=\frac{180-C}{2} = \frac{\angle{BPQ}}{2}.\]Hence, $PI$ bisects $\angle{BPQ}$ and we can similarly get that $QI$ bisects $\angle{PQC}$. Thus, the $A$-excircle of $APQ$ must be the incircle of $ABC$, meaning $PQ$ is tangent to the incircle of $ABC$, as desired.

Dumb me did this problem using Phantoms points Let $X$ be a point on the incircle such that $PX$ is tangent to the incircle. Then define $Q'$ as $PX \cap \odot(CIQ)$ and then just angle chase and show that $\angle Q'QC=0$ (also use the fact that points $B,C,P,Q$ are concyclic)