KöMaL 2020 B5071

Let us consider a triangle $ABC$. Let $A_1$ and $A_2$ denote the points that divide side $BC$ into three equal parts, with $A_1$ lying closer to vertex $B$. Analogously, let $B_1$, $B_2$ divide side $CA$ into equal parts, with $B_1$ lying closer to $C$, and let $C_1$ és $C_2$ divide $AB$ into three equal parts, with $C_1$ lying closer to $B$. Prove that the triangles $A_1B_1C_1$ and $B_2C_2A_2$ are congruent, and both of their areas are equal to the third of the area of the triangle $ABC$.