Question 95
Given triangle $ABC$ with $AB = c$, $BC = a$, $CA = b$. The pairs of points $C_1$ and $C_2$, $A_1$ and $A_2$, $B_1$ and $B_2$ are marked on the sides $AB$, $BC$, $CA$, respectively so that the following equalities are valid:
$$
\begin{aligned}
\frac{CA_1}{a} &= \frac{CB_2}{b} = \frac{a+b}{a+b+c}, \\
\frac{AB_1}{b} &= \frac{AC_2}{c} = \frac{b+c}{a+b+c}, \\
\frac{BC_1}{c} &= \frac{BA_2}{a} = \frac{a+c}{a+b+c}.
\end{aligned}
$$
Prove that the points of intersection of the lines $A_1C_2$, $C_1B_2$ and $B_1A_2$ belong to the circumcircle of the triangle $ABC$.
$$
\begin{aligned}
\frac{CA_1}{a} &= \frac{CB_2}{b} = \frac{a+b}{a+b+c}, \\
\frac{AB_1}{b} &= \frac{AC_2}{c} = \frac{b+c}{a+b+c}, \\
\frac{BC_1}{c} &= \frac{BA_2}{a} = \frac{a+c}{a+b+c}.
\end{aligned}
$$
Prove that the points of intersection of the lines $A_1C_2$, $C_1B_2$ and $B_1A_2$ belong to the circumcircle of the triangle $ABC$.