Since the lines rotate uniformly the angles between the lines $a_1(t_1), a_1(t_2)$ and between the lines $a_2(t_1), a_2(t_2)$ are equal for any moments $t_1, t_2$. Hence the points $A_1, A_2, A(t_1), A(t_2)$ lie on the circle, i.e. all points $A(t)$ lie on the same circle. Moreover if the lines rotated by $\phi$ then the point $A(t)$ passed along the arc with the measure $2\phi$. Therefore the points $A(t)$, $B(t)$ and $C(t)$ are moving uniformly and with the same angular velocity along some three circles. And one full-circle rotation there were two moments $t_1$ and $t_2$ such that the triangles $A(t_1)B(t_1)C(t_1)$ and $A(t_2)B(t_2)C(t_2)$ were equilateral and equally oriented.

Consider this situation on the complex plane and without loss of generality let the points make one full-circle rotation during one unit of time. Let $o_1, o_2, o_3$ be the coordinates of the centers and $r_1, r_2, r_3$ be the radii of the corresponding circles. Then the points $A(t), B(t)$ and $C(t)$ have the coordinates $o_1 + r_1 e^{t + \varphi_1}$, $o_2 + r_2 e^{t + \varphi_2}$ and $o_3 + r_3 e^{t + \varphi_3}$ where $\varphi_1, \varphi_2$ and $\varphi_3$ are angles corresponding to the initial positions of points. Then we know that
$$\frac{(o_2 + r_2 e^{t_1 + \varphi_2}) - (o_1 + r_1 e^{t_1 + \varphi_1})}{(o_3 + r_3 e^{t_2 + \varphi_2}) - (o_1 + r_1 e^{t_2 + \varphi_1})} = \frac{(o_2 + r_2 e^{t_2 + \varphi_2}) - (o_1 + r_1 e^{t_2 + \varphi_1})}{(o_3 + r_3 e^{t_3 + \varphi_2}) - (o_1 + r_1 e^{t_3 + \varphi_1})} = \xi,$$
where $\xi$ is the third root of unity. Note that after simple transformations the equation
$$\frac{(o_2 + r_2 e^{t+\varphi_2}) - (o_1 + r_1 e^{t+\varphi_1})}{(o_3 + r_3 e^{t+\varphi_3}) - (o_1 + r_1 e^{t+\varphi_1})} = \xi$$
reduces to a linear equation of the variable $e^t$. Since this equation has two different solutions $e^{t_1}$ and $e^{t_2}$ on a unit circle, it must be degenerate, i.e. it is an identity. Therefore the triangle $A(t)B(t)C(t)$ is always equilateral.