See here

Lemma. $ABC$ given and $P, Q$ on $AB, AC$, $T= CP\cap BQ$, $R = AT\cap PQ$. Let $RH$ be the perpendicular from $R$ on $BC$, then $RH$ is angle $AHT$ bissector. Now, in the main problem let $W$ is the circumcenter of $(APQ)$. Let the circle $\omega$ through $B, C$ is tangent to $(APQ)$ at $S'\not= A$. Consider the tangent line to $(ABC)$ at $A$ and intersect it with $BC$ at $L$. From radical center theorem we get that $S'L$ is tangent to $\omega$, $(APQ)$ at $S'$. Let the circle $(AS'LW)$ meet $BC$ at $L, J$. So $OJ\perp BC$. $WA = WS'$ and $AWS'J$ is cyclic, so $JW$ is angle $\angle S'JA$ bisector. Note that $OJ$ is passing through the middle of $PQ$ which coincides with $PQ\cap AO$, so from lemma we get that $J, S', O$ are collinear. From symmetry $A'\in JO$ and $S' = S$. $\Box$

Replace label $O$ with $K$ and let $O$ be the center of $\odot(ABC)$ instead. Let a line through $A$ parallel to $BC$ cut $\omega$ and $\odot(ABC)$ at $X,Y$ resp. Let $T$ be the circumcenter of $\Delta ASA'$, which clearly lies on $BC$. First, note that $\Delta XPQ$ and $\Delta A'CB$ are hmothetic so $A', K, X$ are colinear. Moreover, if we let $A_1$ be the point diametrically opposite to $A$ w.r.t. $\odot(ABC)$. Then, $$\measuredangle ASA' = \measuredangle ASX = \measuredangle AA_1Y = \measuredangle(AO, AA'). $$But $\measuredangle TAA' = \measuredangle ASA' - 90^{\circ}$ so $AT\perp AO$. Hence $TS^2=TA^2 = TB\cdot TC$ and we are done.

Davrbek wrote: On the sides $AB$ and $AC$ of the triangle $ABC$, the points $P$ and $Q$ are chosen, respectively, so that $PQ||BC$. Segments of $BQ$ and $CP$ intersect at point $O$. Point $A'$ is symmetric to point $A$ relative to line$ BC$. The segment $A'O$ intersects circle $w$ circumcircle of the triangle $APQ$, at the point $S$. Prove that circumcircle of $BSC$ is tangent to the circle $w$. Fix $k \in \mathbb{R}$ with $\overrightarrow{BC}=(1+k)\overrightarrow{PQ}$. Redefine $S \ne A$ as the point on $\omega$ with $\odot(BSC)$ tangent to $\omega$. Let tangent at $A$ to $\odot(ABC)$ meet $BC$ at $T$. By radical axis theorem applied to $\odot(BSC), \odot(ABC), \odot(APQ)$ we conclude that $ST$ is tangent to $\omega$. Let $AA_1$ be an altitude in $\triangle ABC$; $S_1$ be the midpoint of $AS$; $O_1$ be the midpoint of $AO$. Now we need to show $A_1,S_1,O_1$ are collinear. Let $Y$ be the circumcenter of $\omega$. Suppose $AA_1MM_1$ is a rectangle. Let $X=AM_1 \cap A_1O_1$. Claim. $\triangle AA_1X \sim \triangle ATY$ (Proof) Let $N$ be the circumcenter of $\triangle ABC$. Then $\triangle AM_1N \sim \triangle AA_1T$ so by spiral similarity, $\triangle ATN \sim \triangle AA_1M_1$ Now $\overrightarrow{AN}=(1+k)\overrightarrow{AY}$. Thus, we need $\overrightarrow{AM_1}=(1+k)\overrightarrow{AX}$ to conclude. Now observe by Van-Aubel's that $\tfrac{AO}{OM}=\tfrac{2}{k}$ hence $\tfrac{AX}{A_1M}=\tfrac{AO_1}{O_1M}=\tfrac{1}{(k+1)}$ which together with $A_1M=AM_1$ and directions proves the claim. $\blacksquare$ Finally, we observe $\angle AA_1X=\angle ATY=\angle ATS_1=\angle AA_1S_1$ as $ATA_1S_1$ is cyclic so $A_1,S_1,O_1$ are collinear.

Let the line parallel to $BC$ through $A$ and $A'O$ intersect at $S'$ and $\omega$ and $T=S'Q\cap BC$. Claim: $APQS'$ is isosceles trapezoid and $A'BS'T$ is parallelogram. Proof: Let $M=A'S\cap BC$. Let's show that $BM=MT$. By Menelaus's and Thales's Theorem \[\frac{TM}{MB}=\frac{OQ}{BO}\times\frac{S'T}{S'Q}=\frac{AP}{AB}\times\frac{AB}{AP}=1 \ (\bigstar)\]$\rightarrow$ $BM=MT$. On the other hand, parallelity of $AS'$ and $BC$ implies that $BC$ bisects the segment $A'S'$ as well. Therefore $MA=\boxed{MA'=MS'} \ (\bigstar\bigstar)$ in the right $A'AS'$ triangle $\rightarrow$ $\boxed{A'BS'T}$ is parallelogram. Combining $(\bigstar)$,$(\bigstar\bigstar)$ and $AS'\parallel BT$ $\rightarrow$ $ABTS'$ and $\boxed{APQS'}$ isosceles trapezoid. Back to main Problem: Let the tangent line to $\omega$ at $A$ and $BC$ intersect at $X$.It is enough to show that $XA=XS$ $\iff$ $X$ is center of circumcircle of triangle $AA'S$ $\iff$ $\frac{\angle AXA'}{2}+\angle ASA'=180^\circ$.But we have already got $\angle SAQ=\angle SS'Q=\angle BA'S'$ using the fact that $TQ\parallel BA'$ (See above). Finally, the last property of angles makes it easy to prove that$\frac{\angle AXA'}{2}+\angle ASA'=180^\circ$ with angle chasing. $\blacksquare$

P',Q' sym P,Q wrt BC,PQ'-P'Q=E by Desargues for BQP'-CPQ': O lies on A'E. (E,EP) cuts AC,AB at H,I then I,H on (EP'B),(EQ'C) but AP.AI=AH.AC so A lies on radical center of (EP'B),(QE'C). Reflect wrt BC so A'E is radical center of (EPB),(EQC) so D lies on 2 circle but by Miquel D lies on (APQ) so D=O. Thus PQO+OCB=PQE=PEB=POB so (APQ) tangent to (BSC)

This solution seems to be new. Solved with Sunaina Pati, Krutarth Shah, and Malay Mahajan. Define $X$ to be the intersection of lines $OA'$ and $BC$ and $M$ be the midpoint of $PQ$. Supose $BC= s \cdot PQ$. By Ceva, we have that $A,M,O$ are collinear. But note that $$\frac{OX}{XA'} = \frac{d(O,BC)}{d(A',BC)} = \frac{d(O,BC)}{d(A,BC)} = \frac{s \cdot d(O,PQ)}{s \cdot d(A,PQ)} = \frac{OM}{MA}$$which means $XM$ and $AA'$ are parallel, giving $XP = XQ$. Now, define $Y$ to be the reflection of $A'$ over $X$. Since $MX$ is parallel to $AA'$, we have that $d(Y,MX) = d(A',MX) = d(A,MX)$. But also, if $D$ is the foot of the $A$-altitude, $AY$ is parallel to $DX$ because of midpoints, and so $AY$ is parallel to $PQ$. These two together imply that $A$ is the reflection of $A$ over the perpendicular bisector of $PQ$ and hence lies on the circumcircle of $\triangle APQ$. This gives that $\angle YSP = \angle YQP = \angle APQ = \angle ABC = \angle PQX$, implying that points $P,S,X,B$ are concyclic. Analogously, $Q,S,X,C$ are concyclic, so $S$ is the miquel point of $P,Q$ and $X$. To finish, we have that $\angle PSB = \angle PXB = \angle PQX = \angle PQS + \angle SQX = \angle PQS + \angle SCB$ so the circumcircles of $BSC$ and $APQ$ are tangent to each other, as desired. $\blacksquare$

Let $K$ be the point on $w$ such that $AK \parallel PQ$ and $T$ lie on $BC$ such that $TA$ is tangent to $(ABC)$. It's easy to see that $(ABC)$ and $w$ are tangent at $A$, so $TA$ is clearly the common tangent of $(ABC)$ and $w$. Consider the homothety taking $QP$ to $BC$. Because $QP \parallel BC$, this homothety must be centered at $O$. Now, observe $$\measuredangle (KP, BC) = \measuredangle KPQ = \measuredangle KAQ = \measuredangle KAC = \measuredangle BCA = \measuredangle A'CB$$which means $KP \parallel A'C$. An analogous process yields $KQ \parallel A'B$. Combining these two results with $QP \parallel BC$, we conclude that the aforementioned homothety maps $KPQ$ to $A'CB$, so $K \in \overline{A'OS}$ follows immediately. Since $AK \parallel \overline{TBC}$ follows from transitivity, symmetry implies $$\measuredangle ATA' = \measuredangle ATB + \measuredangle BTA' = 2 \measuredangle ATB$$$$= 2 \measuredangle TAK = 2 \measuredangle ASK = 2 \measuredangle ASA'.$$Hence, because $TA = TA'$ by symmetry, we know $T$ is the center of $(ASA')$. Now, $TA = TS$ means $TS$ is tangent to $w$ by equal tangents and $$TS^2 = TA^2 = Pow_{(ABC)}(T) = TB \cdot TC$$implying $TS$ is tangent to $(BSC)$, which finishes. $\blacksquare$

always a fun day to do moving points!

Let $T=OS \cap \omega$. Claim: $AT \parallel BC$. If so, let tangent to $\omega$ at $A$ intersect $BC$ at $X$. By easy angle chasing we can get that $X$ is center of $(ASA')$ and so $XS^2=XA^2=XB\cdot XC$ (the last is since $(APQ)$ and $(ABC)$ are tangent) and we are done. Sketch for proof of claim: fix $A, T, P, Q$ and move linearity $B, C$. Then points $O, A'$ also move linearity. We want to prove that $T, O, A'$ are collinear, so it's enough to check 3 cases. 1) $B=P$, $C=Q$, $O$ is a midpoint of $PQ$. 2) $B=A=C=A'=O$. 3) $BC$ is midline of $\Delta APQ$. All cases are obvious (but in last I bashed in complex numbers...).

Let $A''$ be point on $APQ$ such that $AA'' \parallel BC$. Since $A''PQ$ and $A'CB$ are homothetic we have that $A''$ lies on $A'O$. Let $A'O$ meet $BC$ at $T$. Claim $1 :PBTS$ and $QCTS$ are cyclic. Proof $:$ Note that $\angle BTA' = \angle AA''S = \angle SPB$ so $PBTS$ is cyclic. we prove the other part with same approach. Claim $2 :TP = TQ$. Proof $:$ Note that $BC \parallel AA''$ and midpoint of $AA'$ lies on $BC$ so $T$ is midpoint of $A'A''$ and since $\angle A'AA'' = 90$ we have that $T$ lies on perpendicular bisector of $AA'$ and since $AA'QP$ is isosceles trapezoid we have that $T$ lies on perpendicular bisector of $PQ$ as well. Now note that $\angle PSB = \angle PTB = \angle PQT = \angle PQS + \angle SQT = \angle PAS + \angle SCT$ so $BSC$ and $APQ$ are tangent.