Estonian Math Competitions

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.









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Alternative solution.

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.