Let $X$ and $Y$ be intersections of $PB$ and $PC$ with $AC$ and $AB$. First, we have $RS$ is antiparallel to $BC$ in $\angle BPC$, therefore so is $VW$. Then $V,W,C$ and $B$ are concyclic, so $PV\cdot PB=PW\cdot PC$. Then from similar triangles $\triangle PXA$ and $\triangle PVU$ we have $\frac{PV}{PU}=\frac{PX}{PA}$ and analogically $\frac{PW}{PU}=\frac{PY}{PA}$. Dividing these we get $\frac{PV}{PW}=\frac{PX}{PY}$, so $PX\cdot PB=PY\cdot PC$. Therefore $XYBC$ is cyclic, so $\angle ABP=\angle YBX=\angle YCX=\angle PCA$. Finally, let $A'$ be image of $A$ in translation, which moves $P$ to $B$ (and $C$ to $Q$, because $BPCQ$ is parallelogram). Then $\angle A'QB=\angle ACP=\angle PBA=\angle A'AB$. Therefore, $AA'BQ$ is cyclic. Therefore $\angle CAP=\angle QA'B=\angle BAQ$, which is what we wanted. Q.E.D.

[asy][asy] unitsize(3.2cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10pt); /* Positions of pairs X,U */ real x=0.54, u=2/3; pair sipc(pair A=(0,0), pair P, path c=unitcircle,real ext=10) { return IP(L(A,P,-1/ext,ext),c); } pair A=dir(110), B=dir(205), C=dir(-25), X=WP(C--A,x), Y=IP(circumcircle(B,C,X),A--B), P=IP(B--X,C--Y), T=sipc(A,P), S=sipc(B,P), R=sipc(C,P), U=WP(P--T,u), V=extension(B,P,U,U+C-A), W=extension(C,P,U,U+B-A), Q=B+C-P, Ap=A+B-P; filldraw(U--V--W--cycle,palegrey,pathpen); filldraw(A--X--Y--cycle,palegrey,invisible); D(unitcircle,heavygreen); D(circumcircle(B,C,X),red); D(A--B--C--cycle,heavygrey+linewidth(1)); D(Q--A--T,purple); DPA(B--Q--C^^A--Ap--B^^Ap--Q^^X--Y^^R--S); DPA(B--S^^C--R,pathpen+linetype("4 4")+linewidth(1)); /* Angle marks */ DPA(anglemark(Ap,A,B,5)^^anglemark(X,B,A,5)^^anglemark(X,C,Y,5)^^anglemark(Ap,Q,B,5),orange); /* Dots and labels */ D("A",A,dir(A)); D("B",B,dir(B)); D("C",C,dir(C)); D("X",X,dir(75)); D("Y",Y,dir(100)); D("P",P,dir(120)); D("T",T,dir(T)); D("S",S,dir(S)); D("R",R,dir(R)); D("U",U,dir(U)); D("V",V,N); D("W",W,SSE); D("Q",Q,dir(Q)); D("A'",Ap,dir(Ap)); [/asy][/asy] Let $X=BP\cap AC,Y=CP\cap AB$. Then $\triangle AXY,\triangle UVW$ are in perspective, so it follows by Desargue's theorem that $XY\cap VW$ is at infinity: i.e. $XY\parallel VW\parallel RS$. Hence $BCXY$ is cyclic by Reim's theorem. Let $A'$ be such that $APBA',ACQA'$ are parallelograms. Then $\angle A'AB=\angle XBA=\angle XCY=\angle A'QB$, so $AA'BQ$ is cyclic. But then $\angle BAQ=\angle BA'Q=\angle PAC$, as required. Comment. Note that the position of $U$ on $AT$ is irrelevant, as the direction of $VW$ is independent of $U$. If we take $U=A$, then removing some more extraneous fluff gives the following problem: The point $P$ lies inside triangle $ABC$ so that $\angle ABP = \angle PCA$. The point $Q$ is such that $PBQC$ is a parallelogram. Prove that $\angle QAB = \angle CAP$. [asy][asy] unitsize(2.8cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10pt); real x=0.54; pair A=dir(110), B=dir(205), C=dir(-25), X=WP(C--A,x), Y=IP(circumcircle(B,C,X),A--B), P=IP(B--X,C--Y), Q=B+C-P; D(unitcircle,heavygreen); D(A--B--C--cycle,heavygrey+linewidth(1)); D(Q--A--P,purple); D(B--P--C--Q--cycle); /* Angle marks */ DPA(anglemark(P,B,A,5)^^anglemark(A,C,P,5),orange); /* Dots and labels */ D("A",A,dir(A)); D("B",B,dir(B)); D("C",C,dir(C)); D("P",P,dir(130)); D("Q",Q,dir(Q)); [/asy][/asy] Which is precisely BrMO2 2013/2.

Solution with Ceva's Theorem: Let $X\equiv RS\cap AB, Y\equiv RS\cap AC, D\equiv BS\cap AC, E\equiv CR\cap AB$. Note that $\overline{RS}$ is anti-parallel to $\overline{BC},$ so $B,X,Y,C$ are concyclic $\implies \widehat{AS}+\widehat{RB}=$ $\angle AXY=\angle ACB=\widehat{AR}+\widehat{RB}$ $\implies \widehat{AS}=\widehat{AR}\implies\angle BEC=\angle BDC\implies B,E,D,C$ are concyclic. Apply Ceva's Theorem for $P$ WRT $\triangle ABC$ we get \[1=\frac{\sin{\angle BAP}}{\sin{\angle PAC}}\cdot\frac{\sin{\angle ACP}}{\sin{\angle PCB}}\cdot\frac{\sin{\angle PBC}}{\sin{\angle PBA}}=\frac{\sin{\angle BAP}}{\sin{\angle PAC}}\cdot\frac{\sin{\angle PBC}}{\sin{\angle PCB}}.\qquad (1)\]where $\angle ACP=\angle PBA$ follows by $B,E,D,C$ are concyclic. On the other hand, apply Ceva's Theorem for $Q$ WRT $\triangle ABC$ and use the fact that $\angle ABQ=\angle AEC=\angle ADB=\angle ACQ$ we get \[1=\frac{\sin{\angle CAQ}}{\sin{\angle QAB}}\cdot\frac{\sin{\angle ABQ}}{\sin{\angle QBC}}\cdot\frac{\sin{\angle QCB}}{\sin{\angle QCA}}=\frac{\sin{\angle CAQ}}{\sin{\angle QAB}}\cdot\frac{\sin{\angle QCB}}{\sin{\angle QBC}}.\qquad (2)\] Combine $(1)$ and $(2)$ and remember the obvious condition that $BQCP$ is a parallelogram, from $\angle BAQ+\angle PAC=\angle CAQ+\angle QAB=\angle BAC<180^\circ$ we deduce that $AP,AQ$ are isogonal with respect to $\angle BAC,$ as desired. $\blacksquare$ [asy][asy] size(9cm); defaultpen(fontsize(10pt)); pair A,B,C,R,S,M,T,X,Y,E,D,V,W,Q,P,X,Y; draw(unitcircle); A=dir(110); B=dir(-155); C=dir(-25); P=dir(-185)/3; D=4*A/7+3*C/7; E=IP(A--B,circumcircle(B,C,D)); P=IP(B--D,C--E); T=IP(P--dir(P-A)*2+P,unitcircle); S=IP(P--dir(P-B)*2+P,unitcircle); R=IP(P--dir(P-C)*2+P,unitcircle); M=5*T/7+2*P/7; W=extension(C,R,M,dir(A-B)+M); V=extension(B,S,M,dir(A-C)+M); Q=extension(B,B+dir(C-P),C,C+dir(B-P)); X=IP(A--B,R--S); Y=IP(A--C,R--S); draw(A--B--C--cycle); draw(A--T); draw(B--S); draw(C--R); draw(R--S,blue); draw(V--W,blue); draw(D--E,blue+dashed); draw(M--T); draw(M--W); draw(B--Q); draw(C--Q); draw(M--V); draw(circumcircle(B,D,E),red+linetype("4 4")); draw(A--Q); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$T$",T,dir(T)); dot("$S$",S,dir(S)); dot("$R$",R,dir(R)); dot("$P$",P,dir(120)*1.5); dot("$W$",W,dir(90)); dot("$V$",V,dir(90)); dot("$D$",D,dir(-90)*1.5); dot("$E$",E,dir(-240)); dot("$U$",M,dir(-135)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$Q$",Q,dir(Q)); [/asy][/asy] Remark: It seems quite easy and similar to All Rusia Olympiad 2011/8. Edit: All right, by a bit of angle-chasing we get $\angle ABP=\angle ACP,$ and then we just use same method in ARO 2011/8 to finish the rest: liberator wrote: The point $P$ lies inside triangle $ABC$ so that $\angle ABP = \angle PCA$. The point $Q$ is such that $PBQC$ is a parallelogram. Prove that $\angle QAB = \angle CAP$.

Isn't this way too easy even for a P1 or I am missing something ? (Took me <10 minutes to solve using geogebra ...) (also I don't know if the proof of the last claim is correct or not) Let $V_U$ , $W_U$ denote the points $V, W$ as a function of $U$ for some $U \in \overline{PT}$. Claim For any two $U_1, U_2 \in \overline{AT}$, $V_{U_1}W_{U_1}||V_{U_2}W_{U_2}$ Proof : Observe that $\angle V_{U_1}U_1P = \angle PAE = \angle V_{U_2}U_2P$, and $\angle U_1PV_{U_1} = \angle V_{U_1}PT = \angle U_2PV_{U_2}$, so $\Delta U_1PV_{U_1} \sim U_2PV_{U_2}$. Similarly $\Delta U_1PW_{U_1} \sim \Delta U_2PW_{U_2}$, and so $\Delta PV_{U_1}W_{U_1} \sim \Delta PV_{U_2}W_{U_2}$. The conclusion follows. $\blacksquare$ So we can WLOG set $U = A$. Let $BP, CP$ hit $AC, AB$ at $E, D$ respectively. Then $\ell_{VW} = \ell_{DE}$ Claim: $RS || DE \Leftrightarrow (B,C,D,E) \text{ is cyclic}$ Proof: $\angle CRS = \angle SBC = \angle EBC$, so $PS || DE \Leftrightarrow \angle CRS = \angle CDE \Leftrightarrow \angle CBE = \angle CDE \Leftrightarrow (B,C,D,E) \text{cyclic}$ $\blacksquare$ Claim: $(B,C,D,E) \text{is cyclic} \Rightarrow \angle PAC = \angle BAQ$. Proof Let $\ell_1$ be the line through $A$ parallel to $DC$, and $\ell_2$ be the line through $A$ parallel to $BE$. Let $P^{\infty}_1, P^{\infty}_2$ be the points at infinity on $\ell_1, \ell_2$ respectively. Note that $P = BP^{\infty}_1 \cap CP^{\infty}_2$ and $Q = BP^{\infty}_2 \cap CP^{\infty}_1$. Then since $ \angle P^{\infty}_2AB = \angle DBE = \angle DCE = \angle P^{\infty}_1AC$, so the two pair of lines $(AP^{\infty}_1, AP^{\infty}_2), (AB, AC)$ are isogonal w.r.t $(AB, AC)$. By the forgotten isogonality lemma, thus $AP, AQ$ are isogonal, as desired (?) $\blacksquare$

liberator wrote: [asy][asy] unitsize(3.2cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10pt); /* Positions of pairs X,U */ real x=0.54, u=2/3; pair sipc(pair A=(0,0), pair P, path c=unitcircle,real ext=10) { return IP(L(A,P,-1/ext,ext),c); } pair A=dir(110), B=dir(205), C=dir(-25), X=WP(C--A,x), Y=IP(circumcircle(B,C,X),A--B), P=IP(B--X,C--Y), T=sipc(A,P), S=sipc(B,P), R=sipc(C,P), U=WP(P--T,u), V=extension(B,P,U,U+C-A), W=extension(C,P,U,U+B-A), Q=B+C-P, Ap=A+B-P; filldraw(U--V--W--cycle,palegrey,pathpen); filldraw(A--X--Y--cycle,palegrey,invisible); D(unitcircle,heavygreen); D(circumcircle(B,C,X),red); D(A--B--C--cycle,heavygrey+linewidth(1)); D(Q--A--T,purple); DPA(B--Q--C^^A--Ap--B^^Ap--Q^^X--Y^^R--S); DPA(B--S^^C--R,pathpen+linetype("4 4")+linewidth(1)); /* Angle marks */ DPA(anglemark(Ap,A,B,5)^^anglemark(X,B,A,5)^^anglemark(X,C,Y,5)^^anglemark(Ap,Q,B,5),orange); /* Dots and labels */ D("A",A,dir(A)); D("B",B,dir(B)); D("C",C,dir(C)); D("X",X,dir(75)); D("Y",Y,dir(100)); D("P",P,dir(120)); D("T",T,dir(T)); D("S",S,dir(S)); D("R",R,dir(R)); D("U",U,dir(U)); D("V",V,N); D("W",W,SSE); D("Q",Q,dir(Q)); D("A'",Ap,dir(Ap)); [/asy][/asy] Let $X=BP\cap AC,Y=CP\cap AB$. Then $\triangle AXY,\triangle UVW$ are in perspective, so it follows by Desargue's theorem that $XY\cap VW$ is at infinity: i.e. $XY\parallel VW\parallel RS$. Hence $BCXY$ is cyclic by Reim's theorem. Let $A'$ be such that $APBA',ACQA'$ are parallelograms. Then $\angle A'AB=\angle XBA=\angle XCY=\angle A'QB$, so $AA'BQ$ is cyclic. But then $\angle BAQ=\angle BA'Q=\angle PAC$, as required. Comment. Note that the position of $U$ on $AT$ is irrelevant, as the direction of $VW$ is independent of $U$. If we take $U=A$, then removing some more extraneous fluff gives the following problem: The point $P$ lies inside triangle $ABC$ so that $\angle ABP = \angle PCA$. The point $Q$ is such that $PBQC$ is a parallelogram. Prove that $\angle QAB = \angle CAP$. [asy][asy] unitsize(2.8cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10pt); real x=0.54; pair A=dir(110), B=dir(205), C=dir(-25), X=WP(C--A,x), Y=IP(circumcircle(B,C,X),A--B), P=IP(B--X,C--Y), Q=B+C-P; D(unitcircle,heavygreen); D(A--B--C--cycle,heavygrey+linewidth(1)); D(Q--A--P,purple); D(B--P--C--Q--cycle); /* Angle marks */ DPA(anglemark(P,B,A,5)^^anglemark(A,C,P,5),orange); /* Dots and labels */ D("A",A,dir(A)); D("B",B,dir(B)); D("C",C,dir(C)); D("P",P,dir(130)); D("Q",Q,dir(Q)); [/asy][/asy] Which is precisely BrMO2 2013/2. Just nice....

It's clear that the choice of $U$ is irrelevant, so we may choose $U=A$, so $V=BS\cap AC$ and $W=CR\cap AB$. Claim: The condition of the problem implies $\angle ABP=\angle ACP$. Proof of Claim: We see that $RS$ is anti-parallel to $BC$ and since $RS\parallel WV$, we have $WV$ antiparallel to $BC$, so $WVCB$ cyclic. Thus, $\angle WBV=\angle WCV$, or $\angle ABP=\angle ACP$. $\blacksquare$ Let $D,E\in(ABC)$ so that $AD\parallel BP$ and $AE\parallel CP$. By DDIT on $PCQB$ with $A$, we see that there is an involution swapping $(AB,AC)$, $(AP,AQ)$, and $(AD,AE)$. We see that \[\angle DAB=\angle ABP=\angle ACP=\angle EAC,\]so $AD$ and $AE$ are isogonal. Clearly $AB$ and $AC$ are isogonal, so this involution is isogonality, so $AP$ and $AQ$ are isogonal, as desired. $\blacksquare$

Note that, by Reim's theorem, $BVWC$ is cyclic. Now, let $BP$ and $CP$ hit $AC$, $AB$ at $E$, $F$ respectively. $PVU$ and $PEA$ are similar, so $PE/PV=PU/PA$, and $PFA,PWU$ are also similar, implying that $PF/PW=PV/PE$. Then, by Power of a Point, $BFEC$ is cyclic. So, naturally redefine point $P$ to be $BE\cap CF$, where $BFEC$ is cyclic. Now, we will prove that $AP$ and $AQ$ are isogonal with barycentric coordinates. They are reflections about $M$, the midpoint of $BC$, so if the homogenized coordinates of $P$ are $(x_P,y_P,z_P)$, then $Q$ is $(-x_P,1-y_P,1-z_P)$. So, we wish to show that $\frac{y_P(1-y_P)}{z_P(1-z_P)}=\frac{b^2}{c^2}$ Letting $E=(e,0,1-e)$ and $F=(f,1-f,0)$, the circumcircle of $(BCE)$ has equation $-a^2yz-b^2xz-c^2xy+(x+y+z)b^2(1-e)x=0$, which means $f=1-\frac{b^2}{c^2}(1-e)$. $BE$ and $CF$ are cevians, which makes it easy to find $P$. Ater homogenization, it is $\left(\frac{ef}{e+f-ef},\frac{e-ef}{e+f-ef},\frac{f-ef}{e+f-ef}\right)$. So, $$\frac{y_P(1-y_P)}{z_P(1-z_P)}=\frac{ef(1-f)}{ef(1-e)}=\frac{1-f}{1-e}=\frac{b^2}{c^2}$$as desired.

Let $E, F$ be the infinite point lie on $AC, AB$. Note that $W=SP \cap UE$, $V=RP \cap UF$, and $\infty_{VW} = EF\cap RS$, by Desargues's theorem, we hence that $\triangle PSR$ and $\triangle UEF$ are perspective. Let $UP, ES, FR$ concurrence at point $T$. Denote $K=AC \cap PS$, $L= AB \cap PR$. While $PSTR \sim PKAL$, conclude that $KL \parallel RS$. By Reim's theorem, $BCKL$ is cyclic. Now $\measuredangle ACP = \measuredangle PBA$, hence that $A\infty_{CP}$ and $A\infty_{BP}$ are isogonal conjugate WRT $\angle BAC$. By Desargues's Involution with quadrilateral $ \{ BP, PC, CQ, QB \} $ and point $A$, hence that $AP$ and $AQ$ are isogonal conjugate WRT $\angle BAC$, so we're done.

Taiwan 2014 TST2, Problem 6 wrote: Let $P$ be a point inside triangle $ABC$, and suppose lines $AP$, $BP$, $CP$ meet the circumcircle again at $T$, $S$, $R$ (here $T \neq A$, $S \neq B$, $R \neq C$). Let $U$ be any point in the interior of $PT$. A line through $U$ parallel to $AB$ meets $CR$ at $W$, and the line through $U$ parallel to $AC$ meets $BS$ again at $V$. Finally, the line through $B$ parallel to $CP$ and the line through $C$ parallel to $BP$ intersect at point $Q$. Given that $RS$ and $VW$ are parallel, prove that $\angle CAP = \angle BAQ$. Solution. Let $X=AC\cap BS$, and $Y=AB\cap CR$. We need to prove that $AP$, and $AQ$ are isogonal w.r.t $\angle BAC$, but since $BPCQ$ is a parallelogram by the Parallelogram Isogonality Lemma it is enough to show that $\angle ABP=\angle ACP$, or $BYXC$ cyclic. But by Desargue's Theorem on $\triangle AXY,\triangle UVW\implies XY\parallel VW\parallel RS$. Now By the converse of Reim's Theorem $BYXC$ cyclic.$\blacksquare$

It suffices to show $\angle ABP = \angle ACP$, as then the result follows by the Parallelogram Isogonality Lemma. (In particular, we may ignore the point $Q$ from the diagram.) Let $X=BP\cap AC$ and $Y=CP\cap AB$. Since $AX \parallel WU$ and $AY \parallel VU$, by Thales' theorem (or the homothety centered at $P$ which maps $A$ to $U$) we obtain $XY \parallel VW$. Now the problem condition $RS \parallel VW$ implies $XY \parallel RS$. Thus $\angle BYX = \angle BSR = \angle CBR = \angle BCX$, i.e. $BCXY$ is cyclic and so $\angle ABP = \angle XBY = \angle XCY = \angle ACP$, as desired.

By the First Isogonality Lemma it suffices to show $\angle ABP = \angle ACP$. Let $W' = \overline{AB} \cap \overline{CP}$ and $V' = \overline{AC} \cap \overline{BP}$. Then we wish to show $BW'V'C$ is cyclic. Note that from Reim's we find that $BRSC$ cyclic implies $BVWC$ cyclic. Thus it suffices to show that $\overline{W'V'} \parallel \overline{WV}$ and we may finish by Reim's. To see this consider the negative homothety $\mathcal{H}$ at $P$ mapping $U \mapsto A$. Consider the image of $\triangle UVW$ under $\mathcal{H}$. Clearly $V \mapsto V'$ and $W \mapsto W'$ due to the condition $\overline{UV} \parallel \overline{AV'}$ and $\overline{UW} \parallel \overline{AW'}$. However then we must have $\overline{VW} \parallel \overline{V'W'}$ so we are done. $\square$