Question 20
An acute-angled triangle $ABC$ is given. The points $A_1$, $A_2$, $B_1$, $B_2$, $C_1$ and $C_2$ lie on its sides in such a way that $\overline{AA_1} = \overline{A_1A_2} = \overline{A_2B} = \frac{1}{3}\overline{AB}$, $\overline{BB_1} = \overline{B_1B_2} = \overline{B_2C} = \frac{1}{3}\overline{BC}$ and $\overline{CC_1} = \overline{C_1C_2} = \overline{C_2A} = \frac{1}{3}\overline{CA}$. Let $k_A$, $k_B$ and $k_C$ be the circumscribed circles of the triangles $AA_1C_2$, $BB_1A_2$ and $CC_1B_2$ respectively, $a_B$ and $a_C$ be the tangents of $k_A$ in $A_1$ and $C_2$, $b_C$ and $b_A$ be the tangents of $k_B$ in $B_1$ and $A_2$ and $c_A$ and $c_B$ be the tangents of $k_C$ in $C_1$ and $B_2$. Prove that the perpendiculars drawn from the intersection of $a_B$ and $b_A$ to $AB$, the intersection of $b_C$ and $c_B$ to $BC$ and the intersection of $c_A$ and $a_C$ to $CA$ intersect in one point.
