BALTIC-WAY 1991 Problem 19

Let's expand a little bit three circles, touching each other externally, so that three pairs of intersection points appear. Denote by $A_{1}, B_{1}, C_{1}$ the three so obtained "external" points and by $A_{2}, B_{2}, C_{2}$ the corresponding "internal" points. Prove the equality

$$\left|A_{1} B_{2}\right| \cdot\left|B_{1} C_{2}\right| \cdot\left|C_{1} A_{2}\right|=\left|A_{1} C_{2}\right| \cdot\left|C_{1} B_{2}\right| \cdot\left|B_{1} A_{2}\right| .$$