Lemma: Let $a,b,c,d \in [0,\frac{\pi}{2}]$ such that: $\frac{sina}{sinb}=\frac{sinc}{sind}$ and $a+b=c+d$.Then,we have: $a=c$ and $b=d$.

Proof Firstly,suppose $(B{P}_A{P}_{C}) \cap (C{P}_{A}{P}_{B}) = S \neq {P}_{A}$. Then,we have: $(S{P}_{B},S{P}_{C})\equiv(S{P}_{B},S{P}_{A})+(S{P}_{A},S{P}_{C})\equiv(CA,CB)+(BC,BA)\equiv(A{P}_{B},A{P}_{C})$ (mod $\pi$) $\implies$ ${S}\in{(A{P}_{B}{P}_{C})}$ Now,let $R$ be the intersection of $BB'$ and $CC'$; $CC' \cap (ABC) = E \neq C$,$BB" \cap (ABC) = F \neq B$ Since $(P P_B,P P_C) \equiv (PB,PC) \equiv (A P_B,A P_C)$ (mod $\pi$) $\implies$ $P \in (A P_B P_C)$ We have $A X_A$,$C X_C$,$S P_B$ are 3 radical axes of $(ABC)$,$(A P_B P_C)$ and $(C P_A P_B)$ $\implies$ $A X_A$,$C X_C$,$S P_B$ are concurrent or $B' \in S P_B$ Same as above, we get $C' \in S P_C$ Thus, it's simple to see that $S P_B FB$ and $S P_C EC$ are cyclic. Let $( C_1 )$ and $( C_2 )$ be $(SFB)$ and $(SEC)$ respectively and let $T$ be the intersection of $( C_1 )$ and $( C_2 )$. We have: $(ER,ES)\equiv(EC,ES)\equiv( P_C C, P_C S)\equiv ( P_C P, P_C S)$ (mod $\pi$) $(FR,FS)\equiv(FB,FS)\equiv( P_B B, P_B S)\equiv ( P_B P, P_B S)$ (mod $\pi$) $( P_C P, P_C S)\equiv( P_B P, P_B S)$ (mod $\pi$) $\implies$ $ERSF$ is cyclic. Because $\overline{RE}.\overline{RC}=\overline{RF}.\overline{RB}$, we get $SR$ is the radical axis of $( C_1 )$ and $( C_2 )$ $\implies$ $\overline{RE}.\overline{RC}=\overline{RF}.\overline{RB}=\overline{RS}.\overline{RT}=k$ and $S$,$R$,$T$ are collinear. Thus, through the agency of ${I}^{k}_R$: $S <-> T$,$E <-> C$,$F <-> B$ Besides,we have $ERSF$ is cyclic,this implies $T$,$B$,$C$ are collinear. Therefore, we get: $(SR,S{P}_{B})\equiv(ST,S{P}_{B})\equiv(BP,BC)$ (mod $\pi$) $(SR,S{P}_{C})\equiv(ST,S{P}_{C})\equiv(CB,CP)$ (mod $\pi$) Set $|(SR,S{P}_{B})|=a$ and $|(SR,S{P}_{C})|=b$ and we can assume $a,b \in [0,\frac{\pi}{2}]$.Moreover,we have $a+b=\widehat{BAC}$ Then,by Sine theorem,we get: $\frac{sina}{sinb}=\frac{PC}{PB}$ $(1)$ Next,set $|(AR,A{P}_{B})|=c$ and $|AR,A{P}_{C}|=d$ $\implies$ $c+d=\widehat{BAC}$ Then,by Sine theorem and some simple calculations,we have: $\frac{sinc}{sind}=\frac{RC}{RB}\cdot\frac{AE}{AF}=\frac{CF}{BE}\cdot\frac{AE}{AF}=\frac{CF}{AF}\cdot\frac{AE}{BE}=\frac{X_C B}{X_A B}\cdot\frac{X_A C}{X_B C}=\frac{X_A C}{X_A B}\cdot\frac{X_C B}{X_B C}=\frac{P_B C}{P_C B}\cdot\frac{X_C B}{X_B C}=\frac{PC}{PB}$ $(2)$ $(1)+(2)$ $\implies$ $\frac{sina}{sinb}=\frac{sinc}{sind}$ $\implies$ According to lemma,we get $a=c$ and $b=d$,this implies $R \in (A P_B P_C)$.Then,we're done!.

[asy][asy] import olympiad; size(10cm); defaultpen(fontsize(9pt)); pen mydash = linetype(new real[] {5,5}); pair A = dir(120); pair B = dir(210); pair C = dir(330); pair P1 = dir(280); pair P = 2*foot(P1, B, C) - P1; pair PA = extension(A, P, B, C); pair PB = extension(B, P, C, A); pair PC = extension(C, P, A, B); pair O = circumcenter(A, B, C); pair OA = circumcenter(A, PB, PC); pair OB = circumcenter(B, PC, PA); pair OC = circumcenter(C, PA, PB); pair XA = 2*foot(A, O, OA) - A; pair XB = 2*foot(B, O, OB) - B; pair XC = 2*foot(C, O, OC) - C; pair A1 = extension(B, XB, C, XC); pair B1 = extension(C, XC, A, XA); pair C1 = extension(A, XA, B, XB); pair Q = 2*foot(PA, OB, OC) - PA; pair K = extension(B, B1, C, C1); pair D = extension(A, A1, B, C); pair E = 2*foot(O, A, D) - A; draw(A--B--C--cycle, black+1); draw(A1--B1--C1--cycle); draw(circumcircle(A, B, C)); draw(circle(OA, abs(OA-A)), royalblue); draw(circle(OB, abs(OB-B)), royalblue); draw(circle(OC, abs(OC-C)), royalblue); draw(circumcircle(K, D, C), orange); draw(A--A1); draw(B--B1, mydash); draw(C--C1, mydash); draw(A--PA); draw(B--PB); draw(C--PC); draw(Q--A1, dotted); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$P_A$", PA, dir(270)); dot("$P_B$", PB, dir(45)); dot("$P_C$", PC, dir(135)); dot("$A'$", A1, dir(270)); dot("$B'$", B1, dir(45)); dot("$C'$", C1, dir(135)); dot("$P$", P, dir(260)); dot("$X_A$", XA, dir(45)); dot("$X_B$", XB, dir(225)); dot("$X_C$", XC, dir(45)); dot("$D$", D, dir(225)); dot("$K$", K, dir(70)); dot("$Q$", Q, dir(330)); dot("$E$", E, dir(225)); [/asy][/asy] Let $\omega_A$, $\omega_B$ and $\omega_C$ be circumcircles of $\triangle AP_BP_C$, $\triangle BP_CP_A$ and $\triangle CP_AP_B$. By Miquel's theorem, these three circles intersect at a point $Q$. Obviously, $A$, $P$, $P_B$, $P_C$ are concyclic. Let $D$ be the Miquel point of this quadrilateral, which lies on $\overline{BC}$. Suppose that $\overline{AD}$ intersects $(ABC)$ at $E$. Claim: $A$, $D$, $A'$ are collinear. Proof. By radical axes on $\omega_B$, $\omega_C$ and $(ABC)$, it follows that $\overline{QP_A}$ passes through $A'$. Therefore, to show the claim, it suffices to show that $A$, $Q$, $P_A$, $E$ are concyclic. Now consider a $\sqrt{bc}$ inversion at $A$. [asy][asy] import olympiad; size(10cm); defaultpen(fontsize(9pt)); pen mydash = linetype(new real[] {5,5}); pair A = dir(120); pair PB = dir(210); pair PC = dir(330); pair P1 = dir(280); pair D = 2*foot(P1, PB, PC) - P1; pair B = extension(PB, D, A, PC); pair C = extension(PC, D, A, PB); pair E = extension(A, D, B, C); pair P = foot(E, PB, PC); pair OA = circumcenter(A, B, C); pair PA = 2*foot(OA, A, P) - A; pair OB = circumcenter(PB, PA, C); pair OC = circumcenter(PC, PA, B); pair Q = 2*foot(OB, PB, PC) - PB; pair F = 2*foot(OA, Q, PA) - PA; draw(circumcircle(A, B, C)); draw(PB--A--PC); draw(PB--PC, royalblue); draw(circle(OB, abs(PB-OB)), royalblue); draw(circle(OC, abs(PC-OC)), royalblue); draw(PB--B); draw(PC--C); draw(B--C); draw(A--D); draw(P--F); draw(A--F); draw(A--P); draw(A--B--D--C--cycle, mediumred+1); draw(Q--F, mydash); dot("$A$", A, dir(A)); dot("$B$", B, dir(350)); dot("$C$", C, dir(130)); dot("$D$", D, dir(280)); dot("$E$", E, dir(270)); dot("$F$", F, dir(45)); dot("$P_B$", PB, dir(225)); dot("$P_C$", PC, dir(315)); dot("$P$", P, dir(270)); dot("$P_A$", PA, 3*dir(265)); dot("$Q$", Q, dir(300)); [/asy][/asy] Notice that $P$ is the Miquel point of cyclic quadrilateral $ACDB$ whose diagonals intersect at $E$. Therefore, by Miquel point properties, $EP \perp BC$ and $\overline{BC}$ is the external angle bisector of $\angle DPA$. Since $\measuredangle BDP = \measuredangle BP_CP = \measuredangle BP_AQ$, we see that $\overline{DP}$ and $\overline{QP_A}$ intersect on $(ABC)$, say at $F$. As $AF \parallel BC$ by Reim's and $\overline{BC}$ bisects $\angle APF$, it follows that $AP_ADF$ is an isosceles trapezoid, so $\overline{PE}$ is its axis of symmetry. Consequently, we see that $E$ lies on $\overline{QP_AF}$ which finishes our proof of the claim. $\square$ Let $\overline{AD}$ intersect $\omega_A$ at $K$. Our goal now is to show that $K$ lies on both $\overline{BB'}$ and $\overline{CC'}$. Angle chasing a bit gives \[ \measuredangle KQP_A = \measuredangle KQP_B + P_BQP_A = \measuredangle KAP_B + \measuredangle P_BCP_A = \measuredangle DAC + \measuredangle ACD = \measuredangle KDP_A, \]so $KDP_AQ$ is cyclic. Therefore, $A'K \cdot A'D = A'P_A \cdot A'Q = A'C \cdot A'X_C$, so $KDCX_C$ is also cyclic. Let its circumcircle be $\omega$. Since $BA \cdot BP_C = BD \cdot BC$ and $B'A \cdot B'X_A = B'C \cdot B'X_C$, it follows that $\overline{BB'}$ is the radical axis of $\omega$ and $\omega_A$, so it passes through $K$. Similarly, $K$ also lies on $\overline{CC'}$ and we are done. $\blacksquare$

Celebrating my 800th post with this cool problem. Israeli Olympic Revenge 2021, Problem 3 wrote: Let $ABC$ be a triangle. A point $P$ is chosen inside $\triangle ABC$ such that $\angle BPC+\angle BAC=180^{\circ}$. The lines $AP,BP,CP$ intersect $BC,CA,AB$ at $P_A,P_B,P_C$ respectively. Let $X_A$ be the second intersection of the circumcircles of $\triangle ABC$ and $\triangle AP_BP_C$ . Similarly define $X_B,X_C$. Let $B'$ be the intersection of lines $AX_A,CX_C$, and let $C$ be the intersection of lines $AX_A,BX_B$. Prove that lines $BB'$ and $CC'$ intersect on the circumcircle of $\triangle AP_BP_C$. [asy][asy] import geometry; import olympiad;size(12cm);defaultpen(fontsize(9pt)); pair A,B,C,P1,P,PA,PB,PC,S,T,A1,B1,C1,XA,XB,XC,R; A=dir(113);B=dir(217);C=dir(333);P1=dir(270); P=2*foot(P1,B,C)-P1;PA=extension(A,P,B,C);PB=extension(B,P,A,C);PC=extension(C,P,A,B); path wa=circumcircle(A,PB,PC);path wb=circumcircle(B,PA,PC);path wc=circumcircle(C,PA,PB);S=intersectionpoints(wa,wb)[1]; path w=circumcircle(A,B,C);XA=intersectionpoints(w,wa)[0];XB=intersectionpoints(w,wb)[1];XC=intersectionpoints(w,wc)[0]; A1=extension(B,XB,C,XC);B1=extension(A,XA,C,XC);C1=extension(B,XB,A,XA);T=extension(B,B1,C,C1);R=extension(T,S,B,C); draw(A--B--C--cycle,red+1);draw(A--PA,orange);draw(B--PB,orange);draw(C--PC,orange);draw(w,heavyblue);draw(wa,heavyblue);draw(wb,heavyblue);draw(wc,heavyblue);draw(A1--B1--C1--cycle,red);draw(A--A1,heavygreen);draw(B--B1,heavygreen);draw(C--C1,heavygreen);draw(S--A1,magenta);draw(S--B1,magenta); draw(S--C1,magenta);draw(T--P,cyan+1); dot("$A$", A, dir(150)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$P$", P, dir(270)); dot("$P_A$", PA, dir(PA)); dot("$P_B$", PB, dir(PB)); dot("$P_C$", PC, dir(PC)); dot("$S$", S, dir(S)); dot("$X_A$", XA, dir(90)); dot("$X_B$", XB, dir(XB)); dot("$X_C$", XC, dir(XC)); dot("$A'$", A1, dir(A1)); dot("$B'$", B1, dir(B1)); dot("$C'$", C1, dir(C1)); dot("$T$", T, dir(T)); [/asy][/asy] Let $A'$ be the intersection of $BX_B$ and $CX_C$. By the angle condition $P$ lies on $(AP_BP_C)$. Let $S$ be the intersection of $(AP_BP_C),(BP_AP_C),(CP_AP_B)$ (concurrency by Miquel theorem). By radical axis, we get that $A'$ lies on $SP_A$, $B'$ lies on $SP_B$ and $C'$ lies on $SP_C$. Furthermore, by radical axis, we get that $(ABC)\cap (SAP_A)$ lies on $AA'$, $(ABC)\cap (SBP_B)$ lies on $BB'$ and $(ABC)\cap (SCP_C)$ lies on $CC'$. Let $T$ be the point on the line connecting $A$ and $(ABC)\cap (ASP_A)$ so that $TP\parallel BC$ and let $R=ST\cap BC$. Let $P'$ denote the reflection of $P$ over $BC$, note that $P'$ lies on $(ABC)$. Claim: $P'$ lies on $(ASP_A)$ (this is only necessary for the second claim). Proof. This is an angle chase. Indeed, we have \begin{align*} \measuredangle P'AS&=\measuredangle P_CAS+\measuredangle P'AP\\&=\measuredangle P_CAS+\measuredangle BCP\\&=\measuredangle BP_CS+\measuredangle P_CSA+\measuredangle BP_AP+\measuredangle P_APC\\&=\measuredangle BP_AP+\measuredangle BP_AS\\&=\measuredangle P'P_AS.\quad \square \end{align*}Claim: $T$ lies on $(AP_BSPP_C)$. Proof. \begin{align*} \measuredangle P_CPT&=\measuredangle P_CCB\\&=\measuredangle BCP'\\&=\measuredangle BAP'\\&=\measuredangle P_CAT.\quad \square\end{align*} Note that $\measuredangle P_AAS=\measuredangle PTS=\measuredangle P_ARS$, thus $R$ lies on $(ASP_A)$. Also, $\measuredangle BP_BS=\measuredangle PTS=\measuredangle BRS$, thus $R$ lies on $(BSP_B)$, and, $\measuredangle CP_CS=\measuredangle PTS=\measuredangle CRS$, thus $R$ lies on $(CSP_C)$. By radical axis again on $(ABC), (ASP_A), (BSP_B)$ and on $(ABC), (ASP_A), (CSP_C)$ implies that $T$ lies on the radical axis of $(ABC)$ and $(BSP_B)$, and, on the radical axis of $(ABC)$ and $(CSP_C)$. We conclude that $T$ lies on $BB'$ and on $CC'$, we are done. $\blacksquare$ [asy][asy] import geometry; import olympiad;size(15cm);defaultpen(fontsize(9pt)); pair A,B,C,P1,P,PA,PB,PC,S,T,A1,B1,C1,XA,XB,XC,R; A=dir(113);B=dir(217);C=dir(333);P1=dir(270); P=2*foot(P1,B,C)-P1;PA=extension(A,P,B,C);PB=extension(B,P,A,C);PC=extension(C,P,A,B); path wa=circumcircle(A,PB,PC);path wb=circumcircle(B,PA,PC);path wc=circumcircle(C,PA,PB);S=intersectionpoints(wa,wb)[1]; path w=circumcircle(A,B,C);XA=intersectionpoints(w,wa)[0];XB=intersectionpoints(w,wb)[1];XC=intersectionpoints(w,wc)[0]; A1=extension(B,XB,C,XC);B1=extension(A,XA,C,XC);C1=extension(B,XB,A,XA);T=extension(B,B1,C,C1);R=extension(T,S,B,C); draw(A--B--C--cycle,red+1);draw(A--PA,orange);draw(B--PB,orange);draw(C--PC,orange);draw(w,heavyblue);draw(wa,heavyblue);draw(wb,heavyblue);draw(wc,heavyblue);draw(A1--B1--C1--cycle,red);draw(R--B,red+1);draw(R--S,red+1);draw(A--A1,heavygreen);draw(B--B1,heavygreen);draw(C--C1,heavygreen);draw(S--A1,magenta);draw(S--B1,magenta); draw(S--C1,magenta);draw(circumcircle(A,S,PA),heavyblue+dashed);;draw(circumcircle(B,S,PB),heavyblue+dashed);;draw(circumcircle(C,S,PC),heavyblue+dashed);draw(T--P,cyan+1); clip((2.4,2.5)--(2.4,-2)--(-4.7,-2)--(-4.7,2.5)--cycle); dot("$A$", A, dir(150)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$P$", P, dir(270)); dot("$P_A$", PA, dir(PA)); dot("$P_B$", PB, dir(PB)); dot("$P_C$", PC, dir(PC)); dot("$S$", S, dir(S)); dot("$X_A$", XA, dir(90)); dot("$X_B$", XB, dir(XB)); dot("$X_C$", XC, dir(XC)); dot("$A'$", A1, dir(A1)); dot("$B'$", B1, dir(B1)); dot("$C'$", C1, dir(C1)); dot("$T$", T, dir(T)); dot("$R$", R, dir(R)); [/asy][/asy]