mathuz wrote: Let $ABCD$ is trapezoid with $AB\parallel CD,$ $ \Omega $ is circle passes through $A,B,C,D.$ $\omega $ is circle passes through $C,D$ and intersects with $CA,CB$ at ${\color{red}A_1,B_1}$ respectively. $A_2$ and $B_2$ are the points symmetric to $A_1$ and $B_1$ respectively, with respect to the midpoints of $CA$ and $CB.$ Prove that the points $A,B,A_2,B_2$ are concyclic. Typo corrected in red color. This is proved in the solution of the problem All Russian-2014, Grade 11, day 2, P2.

you are right! Thank you Luis.

$D$ is a center of spiral similar which goes $BB_{1}$ to $AA_{1}$ $\Rightarrow \frac{AA_{1}}{BB_{1}}=\frac{AD}{BD}=\frac{BC}{CA}$ $AA_{1}.AC=BB_{1}.BC\Rightarrow CA_{2}.AC=CB_{2}.BC$ so $A_{2}B_{2}BA$ is cycle. done

$ABCD$ is cyclic in $\Omega$. So, $\measuredangle BAC = \measuredangle DCA \Rightarrow BC = AD$ Similarly $BD = AC$. Now let $\left\{D,D'\right\} = AD \cap \omega$. And by symmetry, $AD' = BB_1$ Now $A_1CDD'$ is cyclic(in $\omega$) and $A_1C \cap DD' = A$. So(using power of point), $AA_1.AC = AD'.AD$. $\therefore CA_2/CB_2=AA_1/BB_1=AA_1/AD'=AD/AC=CB/CA \Rightarrow CA_2.CA=CB_2.CB$ $\therefore A_2, B_2, A, B$ are concyclic. $[QED]$

What a nice illustration of spiral similarity. Though I would have just said "isosceles trapezoid" in the problem statement. [asy][asy] size(8cm); defaultpen(fontsize(9pt)); pair A = dir(110); pair B = dir(70); pair C = dir(-40); pair D = dir(220); pair B_1 = 0.3*C+0.7*B; pair O = circumcenter(B_1, C, D); pair A_1 = -C+2*foot(O, C, A); pair A_2 = A+C-A_1; pair B_2 = B+C-B_1; draw(CP(O, D), blue); draw(unitcircle, blue); draw(A--B--C--D--A--C); draw(circumcircle(A, B, A_2), red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$B_1$", B_1, dir(B_1)); dot("$A_1$", A_1, dir(60)); dot("$A_2$", A_2, dir(-15)); dot("$B_2$", B_2, dir(B_2)); [/asy][/asy] We have $\triangle DAA_1 \sim \triangle DBB_1$, but $DA = CB$ and $DB = AC$. So $AA_1 \cdot AC = BB_1 \cdot BC$, implying that $CA_2 \cdot CA = CB_2 \cdot CB$.

Let $\omega$ intersect $BD$ in $A'$ Then it is easy to see $AA_1=BA'$ But since $A'B_1CD$ are concyclic, $BB_1 \cdot BC = BA' \cdot BD$ $\implies BB_1 \cdot CB=AA_1 \cdot CA$ $\implies CB_2 \cdot CB= CA_2 \cdot CA$ QED

easy!! but still posting! let $\omega$ intersect $AD$ in $K$ than quite easily $\angle AKB_1=\angle ACD=180-\angle ABB_1$ and hence $ABB_1K$ is isosceles trapezium. now by POP we have $AD.AK=AA_1.AC=CA_2.AC$ on other note $AD.AK=BC.AK=BC.BB_1=BC.BB_2$ and hence $BC.BB_2=CA_2.CA$ and hence by POP we have $ABB_2A_2$ is cyclic quad. thus we are done

Another solution: Let the circle $AA_2B$ intersect $AB$ again at $B'$. Now, $AB$ is the radical axis of $(ABCD) ; (AA_2B)$ and $CD$ is the radical axis of $(ABCD) ;(DCA_1B_1)$. Now, $AB \parallel CD$ and so $AB \parallel CD \parallel l$ where $l$ is the radical axis of $(AA_2B) ; (DCA_1B_1)$ Let $M$ and $N$ be the mid points of $CA,CB$ respectively. It is evident that $MN \parallel l$ and also, $MA_1.MC=MA_2.MA$ so $M$ lies on $l$. Therefore, $N$ lies on $l$ too and so by power of a point $B_2 \equiv B'$ thus, the result holds.

v_Enhance wrote: What a nice illustration of spiral similarity. Though I would have just said "isosceles trapezoid" in the problem statement. [asy][asy] size(8cm); defaultpen(fontsize(9pt)); pair A = dir(110); pair B = dir(70); pair C = dir(-40); pair D = dir(220); pair B_1 = 0.3*C+0.7*B; pair O = circumcenter(B_1, C, D); pair A_1 = -C+2*foot(O, C, A); pair A_2 = A+C-A_1; pair B_2 = B+C-B_1; draw(CP(O, D), blue); draw(unitcircle, blue); draw(A--B--C--D--A--C); draw(circumcircle(A, B, A_2), red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$B_1$", B_1, dir(B_1)); dot("$A_1$", A_1, dir(60)); dot("$A_2$", A_2, dir(-15)); dot("$B_2$", B_2, dir(B_2)); [/asy][/asy] We have $\triangle DAA_1 \sim \triangle DBB_1$, but $DA = CB$ and $DB = AC$. So $AA_1 \cdot AC = BB_1 \cdot BC$, implying that $CA_2 \cdot CA = CB_2 \cdot CB$. Can we just use symmetry to prove BB1×CB＝AA1×CA

Nice problem for depicting the strength of power of a point Here is my solution Claim $AA_1\times AC=BB_1\times BC$ Proof It needs to be proven that $XA^2-R^2=XB^2-R^2$ where X is the center of the circle passing through the points $A_1,B_1,C$ and $D$ and R is its radius.Since a circle passes through this trapezoid, it is isosceles, so we get that the perpendicular bisectors of $AB$ and $CD$ coincide implying $XA=XB$. Thus the powers of $A$ and $B$ with respect to this circle are equal! The conclusion follows by just taking the isotomic conjugates.

Since $A$ and $B$ are symmetric with respect to $\omega$, by power of a point we have $$CA_2 \cdot CA=AA_1 \cdot AC=\mathrm{Pow}_\omega (A)=\mathrm{Pow}_\omega (B)=BB_1\cdot BC=CB_2 \cdot CB,$$hence $ABA_2B_2$ is cyclic. $\blacksquare$

What we have: $AA_1=A_2C$, $BB_1=CB_2$ and an isosceles trapezoid. Simple angle chase gives $\triangle DAB\sim \triangle DA_1B_1$ and $D$ is sipiral similarity center which goes $BB_1$ to $AA_1$. Thus $AA_1\cdot DB=BB_1\cdot DA$. Since $AA_1\dot AC=CA_2\cdot AC=BC\cdot CB_2$. Hence $ABA_2B_2$ is cyclic.