Denote $\angle BAC = \alpha$ (Figures 7 and 8 present two possible cases). Note that $AC = AF$ and $BC = BF$ as they are radii of circles $\rho$ and $\sigma$, respectively. Thus the triangles $ABF$ and $ABC$ are equal by three equal sides. Hence $\angle BAF = \alpha$. From equal radii of $\sigma$, we also obtain $BD = BC$. Now by inscribed angles subtending equal chords of the circumcircle of the triangle $ABC$, we get $\angle BAD = \alpha$. Consequently, points $A$, $D$ and $F$ lie on one line.

From the inscribed angles of the circumcircle of the triangle $ABC$, we get $\angle ADC = \angle ABC$. As in Solution 1, we can show that the triangles $ABF$ and $ABC$ are equal; thus $\angle ABC = \angle ABF = \frac{\angle CBF}{2}$. If $D$ and $A$ lie on the same side of the line $CF$ (Fig. 9) then $\frac{\angle CBF}{2} = 180^\circ - \angle CDF$ by the inscribed angle theorem. If $D$ and $A$ lie on different sides of the line $CF$ (Fig. 10) then, analogously, $\frac{\angle CBF}{2} = \angle CDF$. Altogether, we have either $\angle ADC = 180^\circ - \angle CDF$ while $D$ and $A$ being on different sides of $CF$ or $\angle ADC = \angle CDF$ while $D$ and $A$ being on different sides of $CF$. In both cases, the obtained equality implies that $A$, $D$ and $F$ lie on the same line.