1977 USAMO Problems/Problem 2

Problem

$ABC$ and $A'B'C'$ are two triangles in the same plane such that the lines $AA',BB',CC'$ are mutually parallel. Let $[ABC]$ denote the area of triangle $ABC$ with an appropriate $\pm$ sign, etc.; prove that $$3([ABC]+ [A'B'C']) = [AB'C'] + [BC'A'] + [CA'B']+ [A'BC]+[B'CA] + [C'AB].$$

Hint

Let the parallel lines $AA', BB', CC'$ be parallel to the $x-axis$, and choose arbitrary origin. Then we can define $A(x_1, a), A'(x_2, a), B(y_1, b), B'(y_2, b), C(z_1, c), C'(z_2, c),$ and so, by the area of a triangle formula, it suffices to prove an algebraic statement that is readily shown to be true.