Theorem Let $ABCDEF$ be a convex hexagon inscribed in a circle. Diagonals $AD, BE$ and $CF$ intersect at one point if and only if $$\frac{AB}{BC} \cdot \frac{CD}{DE}\cdot \frac{EF}{FA}=1.$$

proof

Let $G\neq C$ be the intersection of line $CC'$ with circle $\omega$. We will prove $$\frac{BG}{DG}\cdot\frac{DJ}{CJ}\cdot\frac{CF}{BF}=1$$so the theorem in convex hexagon $FABGDJC$ finishes the proof. We are going to use similarities given by cyclic quadrilaterals. $$\triangle BGC'\sim \triangle CAC',\ \triangle DGC'\sim \triangle CDC'$$$$\frac{BG}{C'G}=\frac{AC}{AC'}\wedge\frac{DG}{C'G}=\frac{CD}{C'D}\implies\frac{BG}{DG}=\frac{AC}{AC'}\cdot\frac{C'D}{CD}$$$$\triangle DJB'\sim \triangle BDB',\ \triangle CJB'\sim \triangle BAB'$$$$\frac{DJ}{B'J}=\frac{BD}{B'D}\wedge\frac{CJ}{B'J}=\frac{AB}{AB'}\implies\frac{DJ}{CJ}=\frac{BD}{B'D}\cdot\frac{AB'}{AB}$$$$\triangle FCA'\sim \triangle BAA',\ \triangle FBA'\sim \triangle CAA'$$$$\frac{CF}{A'C}=\frac{AB}{AA'}\wedge\frac{BF}{A'B}=\frac{AC}{AA'}\implies \frac{CF}{BF}=\frac{AB}{AC}\cdot\frac{A'C}{A'B}$$Therefore $$\frac{BG}{DG}\cdot\frac{DJ}{CJ}\cdot\frac{CF}{BF}=\frac{C'D\cdot BD\cdot AB'\cdot A'C}{AC'\cdot CD\cdot B'D\cdot A'B}.$$Now multiply the following equalities (similarities by tangent-chord angle theorem) $$\triangle DAC'\sim\triangle BDC'\implies \frac{C'D}{AC'}=\frac{BD}{AD},$$$$\triangle AB'D\sim\triangle DB'C\implies\frac{AB'}{B'D}=\frac{AD}{CD},$$$$\triangle CA'D\sim\triangle DA'B\implies\left(\frac{A'C}{A'D}=\frac{CD}{BD}\wedge \frac{A'D}{A'B}=\frac{CD}{BD}\right),$$yielding $$\frac{C'D\cdot AB'\cdot A'C}{AC'\cdot B'D\cdot A'B}=\frac{CD}{BD}.$$QED

motivation

#1672

To solve this problem, we first prove the following result. Given triangle $ABC$, lines $BC$, $AC$, $AB$ intersect a line $\ell$ at $A'$, $B'$, $C'$, respectively. Let $F$ be a point on $AA'$, $P = AC \cap BF$, $Q = AB \cap CF$, $D = PQ \cap \ell$, and $E = BB' \cap CC'$. Then $D$, $E$, and $F$ are collinear. [asy][asy] unitsize(3 cm); defaultpen(fontsize(10)); pair[] A, B, C; pair D, E, F, G, P, Q; A[0] = (-0.226,0.52); B[0] = (-0.57,0.08); C[0] = (0.3,0); B[1] = interp(A[0],C[0],1.6); C[1] = interp(A[0],B[0],2.34); A[1] = extension(B[0],C[0],B[1],C[1]); F = interp(A[0],A[1],0.233); P = extension(A[0],C[0],B[0],F); Q = extension(A[0],B[0],C[0],F); D = extension(P,Q,B[1],C[1]); E = extension(B[0],B[1],C[0],C[1]); draw(Q--C[1], gray(0.7)); draw(Q--C[0], gray(0.7)); draw(A[0]--B[1], gray(0.7)); draw(A[0]--A[1], gray(0.7)); draw(interp(A[1],C[1],-0.1)--interp(A[1],C[1],1.1), gray(0.7)); draw(B[0]--A[1], gray(0.7)); draw(B[0]--F, gray(0.7)); draw(Q--D, gray(0.7)); draw(B[0]--B[1], gray(0.7)); draw(C[0]--C[1], gray(0.7)); draw(D--F, red + dashed); dot("$A$", A[0], NW); dot("$A'$", A[1], S); dot("$B$", B[0], NW); dot("$B'$", B[1], S); dot("$C$", C[0], NE); dot("$C'$", C[1], S); dot("$D$", D, S); dot("$E$", E, S); dot("$F$", F, NE); dot("$P$", P, SW); dot("$Q$", Q, N); label("$\ell$", interp(A[1],C[1],-0.1), dir(0)); [/asy][/asy] Proof. Let $G = PQ \cap BC$. Then $(B,C;A',G) = -1$. [asy][asy] unitsize(3 cm); defaultpen(fontsize(10)); pair[] A, B, C; pair D, E, F, G, P, Q; A[0] = (-0.226,0.52); B[0] = (-0.57,0.08); C[0] = (0.3,0); B[1] = interp(A[0],C[0],1.6); C[1] = interp(A[0],B[0],2.34); A[1] = extension(B[0],C[0],B[1],C[1]); F = interp(A[0],A[1],0.233); P = extension(A[0],C[0],B[0],F); Q = extension(A[0],B[0],C[0],F); D = extension(P,Q,B[1],C[1]); E = extension(B[0],B[1],C[0],C[1]); G = extension(P,Q,B[0],C[0]); draw(Q--C[1], gray(0.7)); draw(Q--C[0], gray(0.7)); draw(A[0]--B[1], gray(0.7)); draw(A[0]--A[1], gray(0.7)); draw(interp(A[1],C[1],-0.1)--interp(A[1],C[1],1.1), gray(0.7)); draw(B[0]--A[1], gray(0.7)); draw(B[0]--F, gray(0.7)); draw(Q--D, gray(0.7)); draw(B[0]--B[1], gray(0.7)); draw(C[0]--C[1], gray(0.7)); dot("$A$", A[0], NW); dot("$A'$", A[1], S); dot("$B$", B[0], NW); dot("$B'$", B[1], S); dot("$C$", C[0], NE); dot("$C'$", C[1], S); dot("$D$", D, S); dot("$E$", E, S); dot("$F$", F, NE); dot("$G$", G, SW); dot("$P$", P, SW); dot("$Q$", Q, N); label("$\ell$", interp(A[1],C[1],-0.1), dir(0)); [/asy][/asy] Since $A = BC' \cap B'C$ and $E = BB' \cap CC'$, we have that $A$, $G$, and $E$ are collinear. [asy][asy] unitsize(3 cm); defaultpen(fontsize(10)); pair[] A, B, C; pair D, E, F, G, P, Q; A[0] = (-0.226,0.52); B[0] = (-0.57,0.08); C[0] = (0.3,0); B[1] = interp(A[0],C[0],1.6); C[1] = interp(A[0],B[0],2.34); A[1] = extension(B[0],C[0],B[1],C[1]); F = interp(A[0],A[1],0.233); P = extension(A[0],C[0],B[0],F); Q = extension(A[0],B[0],C[0],F); D = extension(P,Q,B[1],C[1]); E = extension(B[0],B[1],C[0],C[1]); G = extension(P,Q,B[0],C[0]); draw(Q--C[1], gray(0.7)); draw(Q--C[0], gray(0.7)); draw(A[0]--B[1], gray(0.7)); draw(A[0]--A[1], gray(0.7)); draw(interp(A[1],C[1],-0.1)--interp(A[1],C[1],1.1), gray(0.7)); draw(B[0]--A[1], gray(0.7)); draw(B[0]--F, gray(0.7)); draw(Q--D, gray(0.7)); draw(B[0]--B[1], gray(0.7)); draw(C[0]--C[1], gray(0.7)); draw(A[0]--E, blue); dot("$A$", A[0], NW); dot("$A'$", A[1], S); dot("$B$", B[0], NW); dot("$B'$", B[1], S); dot("$C$", C[0], NE); dot("$C'$", C[1], S); dot("$D$", D, S); dot("$E$", E, S); dot("$F$", F, NE); dot("$G$", G, SW); dot("$P$", P, SW); dot("$Q$", Q, N); label("$\ell$", interp(A[1],C[1],-0.1), dir(0)); [/asy][/asy] Finally, by applying the dual of Pappus's Theorem to the three lines through $A'$ and the three lines through $Q$, we see that $AE$, $CC'$, and $DF$ are concurrent (at $E$). [asy][asy] unitsize(3 cm); defaultpen(fontsize(10)); pair[] A, B, C; pair D, E, F, G, P, Q; A[0] = (-0.226,0.52); B[0] = (-0.57,0.08); C[0] = (0.3,0); B[1] = interp(A[0],C[0],1.6); C[1] = interp(A[0],B[0],2.34); A[1] = extension(B[0],C[0],B[1],C[1]); F = interp(A[0],A[1],0.233); P = extension(A[0],C[0],B[0],F); Q = extension(A[0],B[0],C[0],F); D = extension(P,Q,B[1],C[1]); E = extension(B[0],B[1],C[0],C[1]); G = extension(P,Q,B[0],C[0]); draw(Q--C[1], gray(0.7)); draw(Q--C[0], gray(0.7)); draw(A[0]--B[1], gray(0.7)); draw(A[0]--A[1], gray(0.7)); draw(interp(A[1],C[1],-0.1)--interp(A[1],C[1],1.1), gray(0.7)); draw(B[0]--A[1], gray(0.7)); draw(B[0]--F, gray(0.7)); draw(Q--D, gray(0.7)); draw(B[0]--B[1], gray(0.7)); draw(C[0]--C[1], heavygreen); draw(A[0]--E, blue); draw(D--F, red); dot("$A$", A[0], NW); dot("$A'$", A[1], S); dot("$B$", B[0], NW); dot("$B'$", B[1], S); dot("$C$", C[0], NE); dot("$C'$", C[1], S); dot("$D$", D, S); dot("$E$", E, S); dot("$F$", F, NE); dot("$G$", G, SW); dot("$P$", P, SW); dot("$Q$", Q, N); label("$\ell$", interp(A[1],C[1],-0.1), dir(0)); [/asy][/asy] Hence, $D$, $E$, and $F$ are collinear. $\blacksquare$ We can then describe how we can apply this result here. As above, let $P = AC \cap BF$ and $Q = AB \cap CF$. (Also, $\ell$ is the tangent at $D$.) Then by Brokard's Theorem, $PQ$ is the polar of $A'$. But $DA'$ is a tangent to the circle, so $D$ also lies on the polar of $A'$. Thus, $D$ lies on $PQ$, so by the result above, $D$, $E$, and $F$ are collinear. [asy][asy] unitsize(2 cm); defaultpen(fontsize(10)); pair[] A, B, C; pair D, E, F, O, P, Q; A[0] = dir(100); B[0] = dir(190); C[0] = dir(340); O = (0,0); D = dir(280); B[1] = extension(D, D + rotate(90)*(D), A[0], C[0]); C[1] = extension(D, D + rotate(90)*(D), A[0], B[0]); A[1] = extension(B[0], C[0], B[1], C[1]); E = extension(B[0],B[1],C[0],C[1]); F = O + reflect(A[0],A[1])*(O) - A[0]; P = extension(A[0],C[0],B[0],F); Q = extension(A[0],B[0],C[0],F); draw(Q--C[1], gray(0.7)); draw(Q--C[0], gray(0.7)); draw(B[0]--F, gray(0.7)); draw(A[0]--A[1], gray(0.7)); draw(A[0]--B[1], gray(0.7)); draw(B[0]--A[1], gray(0.7)); draw(C[1]--A[1], gray(0.7)); draw(B[0]--B[1], gray(0.7)); draw(C[0]--C[1], gray(0.7)); draw(Q--D, gray(0.7)); draw(Circle(O,1)); draw(D--F, red); dot("$A$", A[0], NW); dot("$A'$", A[1], dir(0)); dot("$B$", B[0], W); dot("$B'$", B[1], S); dot("$C$", C[0], NE); dot("$C'$", C[1], SW); dot("$D$", D, S); dot("$E$", E, S); dot("$F$", F, NE); dot("$P$", P, SW); dot("$Q$", Q, N); [/asy][/asy] Note: This problem has also been posted here: https://artofproblemsolving.com/community/c6h1714145p11065986

If you take a homography and coordinate bash this problem, will there be config issues?