Converse Corresponding Angles Theorem

This theorem can be proven by an indirect proof. Let ${\ell }_1$ and ${\ell }_2$ be two lines intersected by a transversal line ${\ell }_3$ forming corresponding congruent angles $\angle 1$ and $\angle 2.$

Since the goal is to prove that ${\ell }_1$ is parallel to ${\ell }_2$, it will be temporarily assumed that ${\ell }_1$ and ${\ell }_2$ are not parallel. By the Parallel Postulate, there exists a line $n$ parallel to ${\ell }_2$ that passes through the point of intersection between ${\ell }_1$ and ${\ell }_3.$ This line forms $\angle 3$ and $\angle 4.$ By the Angle Addition Postulate, $m\angle 1$ is equal to the sum of $m\angle 3$ and $m\angle 4.$ Since $n$ and ${\ell }_2$ are parallel lines that are cut by a transversal, by the Corresponding Angles Theorem, $\angle 3$ and $\angle 2$ are congruent. By the definition of congruence, these angles have the same measure. By the Substitution Property of Equality, $m\angle 2$ can be substituted for $m\angle 3$ into the equation for $m\angle 1.$ From the above equation and since $m\angle 4$ is a positive number, it can be concluded that $m\angle 1$ is greater than $m\angle 2.$ This contradicts the given fact that $\angle 1$ and $\angle 2$ are congruent. The contradiction came from assuming that ${\ell }_1$ and ${\ell }_2$ are not parallel lines. Therefore, ${\ell }_1$ and ${\ell }_2$ must be parallel lines.