This one was note the symmetry and lots of angle chasing.

Solution please?

Can you produce a figure? In my figure, ABPQ don't look cyclic.

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Let $W_1,W_2$ relatively be the center of $\omega _1, \omega _2$. Let $G,F$ be the second intersection of $AB$ and $\omega _1,\omega _2$. Since $XB \parallel DE$ and $CE \parallel AX$, it's easy to prove $CEAX, EDXB,W_1EW_2X$ are the parallelogram. Therefore, $(C,A),(D,B),(E,X),(W_1,W_2)$ are the pairs of symmetric point wrt midpoint $H$ of $W_1W_2$. This implies $CDAB$ is the parallelogram. Therefore $BC=AD$. Also, $QF=AD$ because $QD \parallel AF$, then $BC=QF \Rightarrow \angle{CPB}=\angle{QAB}$ $\Rightarrow ABPQ$ is cyclic.

isomorphism wrote: Can you produce a figure? In my figure, ABPQ don't look cyclic. $B$ must lie on $\omega_2$ and $A$ on $\omega_1$

[asy][asy] size(6cm); pair o1 = MP("O_1", (-5, 0)); pair o2 = MP("O_2", (3, 0)); real r = 6; D(o1 -- o2); pair X = MP("X", IP(D(CR(o1, r)), D(CR(o2, r))), dir(90)); pair E = MP("E", OP(D(CR(o1, r)), D(CR(o2, r)))); D(X -- E); pair M = (X + E)/2; MP("M", M, dir(-45)); [/asy][/asy] The homothety $\mathcal{H}(M, -1)$ sends $\omega_1 \to \omega_2$ and $\omega_2 \to \omega_1$ since $MO_1 = MO_2$ and the radiuses are equal. [asy][asy] size(7cm); pair o1 = (-5, 0); //MP("O_1", (-5, 0)); pair o2 = (3, 0); //MP("O_2", (3, 0)); real r = 6; pair X = D(MP("X", IP(D(CR(o1, r)), D(CR(o2, r))), dir(90))); pair E = D(MP("E", OP(D(CR(o1, r)), D(CR(o2, r))))); D(X -- E); pair M = (X + E)/2; D(MP("M", M, dir(0))); real c_angle = 170; real d_angle = 15; pair C = D(MP("C", r*dir(c_angle) + o1, dir(c_angle))); pair D = D(MP("D", r*dir(d_angle) + o2, dir(d_angle))); pair Q = D(MP("Q", OP(D(C -- D), CR(o1, r)), dir(45))); pair P = D(MP("P", IP(D(C -- D), CR(o2, r)), dir(90 + 45))); real f = -1; pair HP = D(MP("P'",M + f*(P - M), 1.1*dir(-45))); pair HQ = D(MP("Q'",M + f*(Q - M), 1.1*dir(180+45))); pair HC = D(MP("C'",M + f*(C - M), 1.1*dir(-25))); pair HD = D(MP("D'",M + f*(D - M), 1.1*dir(180+25))); pair HX = E; D(C -- X -- D); D(HC -- HX -- HD); D(HC -- HD); D(C -- HC); D(D -- HD); [/asy][/asy] Applying $\mathcal{H}$, sends $\triangle CXD \to \triangle C'ED'$. $C' \in \omega_2$ because $\omega_1 \to \omega_2$. Likewise $D' \in \omega_1$ and $X' = E$. In $\square CXC'E$, $CM = C'M$ and $MX = ME$. Therefore, $\square CXC'E$ is a parallelogram $\implies CX \parallel EC' \implies A = C'$. Likewise, $DX \parallel ED'$ and $B = D'$. [asy][asy] size(8cm); pair o1 = (-5, 0); //MP("O_1", (-5, 0)); pair o2 = (3, 0); //MP("O_2", (3, 0)); real r = 6; pair X = D(MP("X", IP(D(CR(o1, r)), D(CR(o2, r))), dir(90))); pair E = D(MP("E", OP(D(CR(o1, r)), D(CR(o2, r))))); D(X -- E); pair M = (X + E)/2; D(MP("M", M, dir(0))); real c_angle = 170; real d_angle = 15; pair C = D(MP("C", r*dir(c_angle) + o1, dir(c_angle))); pair D = D(MP("D", r*dir(d_angle) + o2, dir(d_angle))); pair Q = D(MP("Q", OP(D(C -- D), CR(o1, r)), dir(45))); pair P = D(MP("P", IP(D(C -- D), CR(o2, r)), dir(90 + 45))); real f = -1; pair HP = D(MP("P'",M + f*(P - M), 1.1*dir(-45))); pair HQ = D(MP("Q'",M + f*(Q - M), 1.1*dir(180+45))); pair HC = D(MP("A",M + f*(C - M), 1.1*dir(-25))); pair HD = D(MP("B",M + f*(D - M), 1.1*dir(180+25))); pair HX = E; pair A = HC; pair B = HD; D(C -- X -- D); D(HC -- HX -- HD); D(HC -- HD); //D(C -- HC); D(D -- HD); D(B -- P); D(Q -- A); D(A -- P); D(B -- Q); D(A -- P -- D); D(A -- HQ -- D); D(A -- B -- Q); MA(A,P,D,2,2,red); MA(red,A,HQ,D,2,2,red); MA(red,A,B,Q,2,2,red); [/asy][/asy] $\square CDC'D'$ or $\square ABCD$ is also a parallelogram and $CD \parallel AB \implies QD \parallel Q'B$. Since $QD = Q'D' = Q'B$ and $QD \parallel Q'B$. Therefore, \[ \angle ABQ = \angle AQ'D = \angle APD = \angle APQ \implies \angle ABQ = \angle APQ \]Therefore, $\square ABPQ$ is cyclic.

a_507_bc wrote: Two circles $\omega_1$ and $\omega_2$ with equal radius intersect at two points $E$ and $X$. Arbitrary points $C, D$ lie on $\omega_1, \omega_2$. Parallel lines to $XC, XD$ from $E$ intersect $\omega_1, \omega_2$ at $A, B$, respectively. Suppose that $CD$ intersect $\omega_1, \omega_2$ again at $P, Q$, respectively. Prove that $ABPQ$ is cyclic. Proposed by Ali Zamani In the definition of $A$ and $B$, the $\omega_1$ and $\omega_2$ should be switched.

Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$ respectively. Since $\omega_1$ and $\omega_2$ have the same radius, we must have $\angle ECX=\angle EAX$. So, $AECX$ and $BEDX$ are parallelograms. Thus, $AE=CX$ and $BE=DX$. Also we have $\angle AEX=\angle EXC$ and $\angle BEX=\angle EXD$. Which implies, $\Delta ABE\cong \Delta CDX\implies \angle BAE=\angle XCD$ and $AB=CD$. $\angle BAC=\angle EAC-\angle BAE=\angle ACX-\angle XCD=\angle ACD\implies AB\parallel CD$. Which means $ABCD$ is a parallelogram. Now, \[PS=BC=AD=QR\implies \Delta O_1PS\cong\Delta O_2QR\implies\angle PO_1S=\angle QO_2R\]\[\implies\angle PBA=\angle PBS=\angle RAQ=\angle AQP\]as desired.

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Just prove that CBAD is a parallelogramme And some angle chase kill the problem