Assume $A_1A_2 = d$, and $AB = a$. Construct an equilateral triangle $A_0B_0C_0$ with side of length $a-d$. Points $A', B', C'$ are chosen on the sides of triangle $A_0B_0C_0$ such that $A'C_0 = A_2C$, $B'A_0 = B_2A$, and $C'B_0 = C_2B$.

Therefore,
$$A'B_0 = a-d-A'C_0 = BC-A_1A_2-A_2C = BA_1$$

$$B'C_0 = B_1C, \quad C'A_0 = C_1A.$$
Since
$$\angle B_1CA_2 = \angle B'C_0A', \quad \angle B_2AC_1 = \angle B'A_0C', \quad \angle C_2BA_1 = \angle C'B_0A',$$
thus
$$\triangle CB_1A_2 \cong \triangle C_0B'A', \quad \triangle AB_2C_1 \cong \triangle A_0B'C', \quad \triangle C_2BA_1 \cong \triangle B_0C'A',$$
which implies that $B'C' = C'A' = A'B' = d$ and triangle $A'B'C'$ is equilateral.
So
$$\angle AB_2C_1 = \angle A_0B'C' = 180^\circ - \angle C'B'A' - \angle A'B'C_0 = 120^\circ - \angle A'B'C_0, \quad \angle C_1B_2B_1 = \angle B_1A_2A_1.$$
In view of $B_2C_1 = B_1B_2 = A_2B_1 = A_1A_2 = d$, triangles $C_1B_2B_1$ and $B_1B_2A_1$ are congruent, implying that $B_1C_1 = A_1B_1$. Together with $C_1C_2 = A_1C_2$, we show that $C_2B_1$ is the perpendicular bisector of $A_1C_1$ and $C_2B_1$ is the height of triangle $A_1B_1C_1$ on side $A_1C_1$. Similarly, $C_1A_2$ and $A_1B_2$ are the altitudes of triangle $A_1B_1C_1$ to the sides $A_1B_1$ and $B_1C_1$ respectively.

Therefore the lines $A_1B_2$, $B_1C_2$ and $C_1A_2$ are concurrent.