Six circles [math]\omega_A[/math], [math]\omega_B[/math], [math]\omega_C[/math], [math]\omega_D[/math], [math]\omega_E[/math], [math]\omega_F[/math], lie tangent around a seventh, [math]\omega_O[/math]. Each is also tangent with the next in sequence, and the intersection points they form with [math]\omega_O[/math], [math]A[/math], [math]B[/math], [math]C[/math], [math]D[/math], [math]E[/math], and [math]F[/math], lie along [math]\omega_O[/math] in that order. In that case, [math]AD[/math], [math]BE[/math], and [math]CF[/math] are concurrent.

Lemma 1: [math]AC[/math], the tangent to [math]\omega_O[/math] at [math]B[/math], and [math]O_AO_C[/math] are all concurrent.[br]Let the second touch points of [math]\omega_B[/math] at [math]\omega_A[/math] and [math]\omega_C[/math] be [math]R[/math] and [math]S[/math], respectively. By angle chasing, [math]ACSR[/math] is cyclic: [br][math]\angle ACS=180^{\circ}-\angle OCA-\angle SCO_C=180^{\circ}-\frac{\left(180^{\circ}-\angle AOC\right)}{2}-\frac{\left(180^{\circ}-\angle CO_CS\right)}{2}=\frac{\angle AOC+\angle CO_CS}{2}[/math][br][br]Similarly we get that [math]\angle SRA=\frac{\left(\angle SO_BR+\angle RO_AA\right)}{2}[/math]. Therefore [math]\angle ACS+\angle SRA=\frac{\angle AOC+\angle CO_CS+\angle SO_BR+\angle RO_AA}{2}=\frac{360^\circ}{2}=180^\circ[/math][br][br]By radical axis theorem on [math]\omega_O[/math], [math]\omega_B[/math], and [math]\left(ACSR\right)[/math], [math]AC[/math], [math]RS[/math], and the tangent to [math]\omega_O[/math] at [math]B[/math] all intersect at the same point. Call this point [math]J[/math], and define [math]K[/math] and [math]L[/math] similarly.[br][br]Let [math]T[/math] and [math]U[/math] be the second intersection of [math]AC[/math] and [math]RS[/math] with [math]\omega_C[/math], respectively.[br]By angle chasing it is easy to show that [math]TU\parallel AR[/math]. This means that [math]\triangle JTU[/math] is similar to [math]\triangle JAR[/math] by homothety at [math]J[/math]. Next, [math]\angle ARO_A=180^{\circ}-\angle SRA-\angle O_BRS=180^{\circ}-\angle JUT-\angle RSO_B=\angle TUS-\angle USO_C=\angle TUS-\angle O_CUS=\angle TUO_C[/math]. This also means [math]\angle O_AAR=\angle O_CTU[/math] since [math]\triangle AO_AR[/math] and [math]\triangle TO_CU[/math] are both isosceles, and so [math]\triangle AO_AR[/math] is similar to [math]\triangle TO_CU[/math]. Therefore the homothety at [math]J[/math] that takes [math]TU[/math] to [math]AR[/math] also takes [math]O_C[/math] to [math]O_A[/math], and it follows [math]O_AO_C[/math] is collinear with [math]J[/math].[br][br]This allows us to redefine [math]J[/math] as the intersection of [math]AC[/math] with [math]O_AO_C[/math]. Now, by Desargues' theorem, we have two triangles in perspective from a point: [math]\triangle ACE[/math], and [math]\triangle O_AO_CO_E[/math]. Therefore we will have that [math]JKL[/math] are collinear.[br][br]By Menalaus' theorem on [math]\triangle ACE[/math] and line [math]JKL[/math], we will have [math]\frac{AJ}{CJ}\cdot\frac{CK}{EK}\cdot\frac{EL}{AL}=1[/math]. We multiply this by [math]\frac{BJ}{BJ}\cdot\frac{DK}{DK}\cdot\frac{FL}{FL}[/math] to get [math]\frac{AJ}{BJ}\cdot\frac{BJ}{CJ}\cdot\frac{CK}{DK}\cdot\frac{DK}{EK}\cdot\frac{EL}{FL}\cdot\frac{FL}{AL}=1[/math]. But this is equal to [math]\frac{AB}{BC}\cdot\frac{AB}{BC}\cdot\frac{CD}{DE}\cdot\frac{CD}{DE}\cdot\frac{EF}{FA}\cdot\frac{EF}{FA}[/math], so we have [math]\left(\frac{AB}{BC}\cdot\frac{CD}{DE}\cdot\frac{DE}{EF}\right)^2=1[/math], which means [math]\frac{AB}{BC}\cdot\frac{CD}{DE}\cdot\frac{EF}{FA}=1[/math].[br][br]Now I claim this condition implies that [math]AD[/math], [math]BE[/math], and [math]CF[/math] all intersect. Let [math]H[/math] be the intersection of [math]AD[/math] and [math]BE[/math], and let [math]F'[/math] be the second intersection of [math]CH[/math] with [math]\omega_O[/math]. Then, we have [math]1=\frac{AH}{EH}\cdot\frac{CH}{AH}\cdot\frac{EH}{CH}=\frac{AB}{DE}\cdot\frac{CD}{F'A}\cdot\frac{EF'}{BC}=\frac{AB}{BC}\cdot\frac{CD}{DE}\cdot\frac{EF'}{F'A}[/math]. This implies that [math]\frac{EF}{FA}=\frac{EF'}{F'A}[/math]. Now there are two possibilities for [math]F[/math], but only one of them will lie on the "right" side of segment [math]EA[/math] (i.e. the segment which does not contain [math]B[/math],[math]C[/math], and [math]D[/math]). Below is a diagram of [math]F[/math] on the wrong side, and you can see that the diagonals don't line up. If we restrict [math]F[/math]'s placement to the right side of [math]EA[/math], then it is fairly clear there is only one point [math]F[/math] on [math]\omega_O[/math] such that [math]\frac{EF}{FA}=\frac{EF'}{F'A}[/math]: if there were 2 valid points [math]F_1[/math] and [math]F_2[/math], then we would WLOG have [math]EF_1[/math] < [math]EF_2[/math], and [math]AF_1[/math] > [math]AF_2[/math] simultaneously, which would imply [math]\frac{EF_1}{AF_1}[/math] < [math]\frac{EF_2}{AF_2}[/math], a contradiction.