51st Ukrainian National Mathematical Olympiad, 3rd Round

Let circles with diameters $B_1{B}_2$, $C_1{C}_2$ touch at point $A$. By analogy we define $B$, $C$ (fig. 22).

Since touching point is a center of homothety that transforms one circle into another, then $A=B_1{C}_2\cap B_2{C}_1$ and so on. Let $D=A_2{C}_1\cap A_1{B}_2$. Points $A$, $B$, $C$, $D$ are cyclic, because $$\angle CDB=\angle B_{2}D{C}_1=\pi -\angle D{B}_2{C}_1-\angle D{C}_1{B}_2=\pi -\angle CAE-\angle BAE=\pi -\angle CAB,$$

$$\begin{gathered} \angle CFA=\angle F{A}_1{C}_2+\angle F{C}_2{A}_1=\angle C{A}_{1}B+\angle F{C}_{2}B=\angle C{A}_{1}B+\angle B{C}_{1}A= \\ =\angle CB{E}_1+\angle E_{1}BA=\angle CBA. \end{gathered}$$

But except point $C$, the line $A_1{B}_2$ intersects the circle at point $D$, hence $A_1{B}_2$, $B_1{C}_2$, $C_1{A}_2$ are concurrent.