My solution: Since $ \angle PBC=\angle PAC=\angle ADS $ , so we get $ B, P, D, R $ are concyclic . Similarly, we can prove $ B, Q, D, S $ are concyclic . Since $ \angle PQS=\angle PQD-\angle SQD=\angle PBC-\angle SBD =\angle PBC-\angle DBC=\angle PBD=\angle PRS $ , so we get $ P, Q, R, S $ are concyclic . Q.E.D

$(PQD)$ is tangent to $RS$ at $D$ by inversion at $D$ which fixes $(ABC)$. Let $T$ be the point on $\overline{RS}$ which also lies on the tangent to $(ABC)$ at $B$. It is well known that $|TB|^2=|TD|^2$, which implies that $T$ lies on the radical axis of $(ABC)$ and $(PQD)$, i.e. $TPQ$ collinear. Now $|TR||TS|=|TB|^2=|TP||TQ|$, and we are done.

Different labeling wrote: Let $ABC$ be a triangle such that the line $BD$ bisects the angle $ABC.$ Let $K,L$ be the points on $AB,AC$ respectively, so that $KL\parallel BC$. The circumcircle of triangle $AKL$ intersects $KD$ and $LD$ in the points $P$ and $Q,$ respectively. Prove that the points $B,P,Q$ and $C$ lie on a common circle. Claim. $BC$ is tangent to $(PQD)$. Proof. $$\measuredangle PDB=\measuredangle DKL=\measuredangle PKL=\measuredangle PQL=\measuredangle PQD,$$the claim follows. Let $R$ be the intersection of tangent from $A$ to $(ABC)$. By homothety, $RA$ is also tangent to $(AKL)$. Also, trivial angle chase yields that $RA=RD$ as $AD$ is the angle bisector of $\angle BAC$. Therefore, $R$ has equal power wrt $(AKL)$ and $(PQD)$. Hence, $R$ lies on $PQ$. Now, $RP\cdot RQ=RA^2=RB\cdot RC$. We are done. [asy][asy] size(7cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair O,A,B,C,D,K,L,P,Q,R; O=(0,0);A=dir(120);B=dir(200);C=dir(340);path w=circumcircle(A,B,C);D=extension(incenter(A,B,C),A,B,C);K=0.45*B+0.55*A; L=extension(A,C,K,foot(K,O,midpoint(B--C)));path g=circumcircle(A,K,L);P=intersectionpoints(K--D,g)[1];Q=intersectionpoints(L--D,g)[1];R=extension(B,C,P,Q); draw(A--B--C--cycle,deep);draw(w,deep);draw(g,deep);draw(D--K,med);draw(D--L,med);draw(R--A,deep);draw(R--Q,light+dashed);draw(R--B,deep); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D));dot("$K$",K,dir(K));dot("$L$",L,dir(L));dot("$Q$",Q,S);dot("$P$",P,dir(P));dot("$R$",R,dir(R)); [/asy][/asy]

SecondWind wrote: $(PQD)$ is tangent to ... Before saying T lies on the radical axis of PQD, shouldn't you first prove that TD is tangent to the circumcircle of PQD?

My sol (not full prove) claim that B,D,P,R concyclic and B,D,Q,S concyclic from angle chasing and we get BD,RP,QS concurrent then use radical lemma

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