Question 24
Points $C_1,C_2$ are chosen on side $AB$ of triangle $ABC$. Similarly, $A_1,A_2$ are on $BC$ and $B_1,B_2$ are on $AC$. The segments $A_1B_2$, $B_1C_2$, $C_1A_2$ have equal lengths, are concurrent, and the angle between any two of them is $60^\circ$. Prove that
$$\frac{A_1A_2}{BC}=\frac{B_1B_2}{CA}=\frac{C_1C_2}{AB}.$$
$$\frac{A_1A_2}{BC}=\frac{B_1B_2}{CA}=\frac{C_1C_2}{AB}.$$