Question 17
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
$$
|A_{1} B_{2}| \cdot |B_{1} C_{2}| \cdot |C_{1} A_{2}| = |A_{1} C_{2}| \cdot |C_{1} B_{2}| \cdot |B_{1} A_{2}|.
$$
$$
|A_{1} B_{2}| \cdot |B_{1} C_{2}| \cdot |C_{1} A_{2}| = |A_{1} C_{2}| \cdot |C_{1} B_{2}| \cdot |B_{1} A_{2}|.
$$