Question 89
On each side of a triangle $ABC$, two points are chosen: points $C_1$ and $C_2$ on side $AB$, points $B_1$ and $B_2$ on $AC$, and points $A_1$ and $A_2$ on $BC$. It happened that the segments $A_1B_2$, $B_1C_2$, and $C_1A_2$ have equal length, meet at one point, and the angle between any two of them is $60^\circ$. Prove that
$$
\frac{A_1A_2}{BC} = \frac{B_1B_2}{CA} = \frac{C_1C_2}{AB}.
$$
(I. Bogdanov)
$$
\frac{A_1A_2}{BC} = \frac{B_1B_2}{CA} = \frac{C_1C_2}{AB}.
$$
(I. Bogdanov)