Suppose that \(\ell, m, n\) are lines, such that \(n\) cuts \(\ell\) and \(m\) at unique points. We are given that \(\beta = \gamma\) and our task is to prove that \(\ell \parallel m.\) Assume, for the sake of contradiction, that \(\ell \not\parallel m.\) Then \(\ell, m, n\) form a triangle:

From this we find:

$$\begin{align}
\alpha + \beta &= \gamma & \tag{I: exterior angles theorem}\\
\alpha + \beta &= \beta & \tag{II: \(\beta = \gamma\) is given} \\
\alpha &= 0^\circ & \tag{III: from II}
\end{align}$$

But if \(\alpha = 0^\circ,\) then \(\ell \cong m.\) This is absurd, because no triangle can have an angle with zero measure. Therefore, it must be that our original assumption was wrong, and thus \(\ell \parallel m\) after all. \(\blacksquare\)