To prove that alternate interior angles are congruent, it will be shown that $\angle 1$ and $\angle 2$ are congruent.

Notice that by definition $\angle 2$ and $\angle 5$ are vertical angles. By the Vertical Angles Theorem, they are therefore congruent angles. Furthermore, by definition $\angle 5$ and $\angle 1$ are corresponding angles. Hence, by the Corresponding Angles Theorem, $\angle 5$ and $\angle 1$ are also congruent angles. Applying the Transitive Property of Congruence, $\angle 2$ and $\angle 1$ can be concluded to be congruent angles as well. The same reasoning applies to the other pair of alternate interior angles. Therefore, when a pair of parallel lines is cut by a transversal, the pairs of alternate interior angles are congruent.

Two-Column Proof

The previous proof can be summarized in the following two-column table.

Apart from the points of intersection, consider two more points on each line.

Next, $A,$ $B,$ $C,$ and $D$ will be translated parallel to the transversal until the points $A,$ $C,$ and $D$ lie on ${\ell }_2.$ Then, $A,$ $B,$ and $C$ will be rotated $180^{\circ }$ about $F.$ It should be noted that since point $D$ lies on the transversal, when translating it to ${\ell }_2$ the point will fall into the same position as $F.$ Therefore, $D$ will not be affected by the rotation around $F.$ After this combination of rigid motions, $A,$ $B,$ $C,$ and $D$ are mapped onto $A^{\prime },$ $B^{\prime },$ $C^{\prime },$ and $D^{\prime }.$ This means that $\angle ADB$ is mapped onto $\angle A^{\prime }{D}^{\prime }{B}^{\prime }.$ Therefore, $\angle ADB$ and $\angle A^{\prime }{D}^{\prime }{B}^{\prime }$ are congruent angles. Note that $D^{\prime }$ and $F$ share the same location. It can also be seen that $A^{\prime }$ lies on $\overrightarrow{FG}$ and $B^{\prime }$ lies on $\overrightarrow{FH}.$ Because of this, $\angle A^{\prime }{D}^{\prime }{B}^{\prime }$ is congruent to $\angle GFH.$ Applying the Transitive Property of Congruence, $\angle ADB$ is congruent to $GFH.$