If a straight line falling on two straight lines make the alternate angles equal to one another, the straight lines will be parallel to one another.
[Desmos graphs can be found here]
Let line EF fall on the straight lines AB and CD with the alternate angles AEF and EFD equal to one another.
We will prove that AB is parallel to CD by showing that they cannot intersect. Suppose that they did intersect at a point G in the direction of B and D.
We will look at triangle GEF. By Proposition 16 (In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles) the exterior angle AEF must be greater than the interior angle EFG. But, by hypothesis, those two angles are equal which produces a contradiction.
Therefore AB and CD do not meet at a point in the direction of B and D. In a similar fashion it can be shown that they do not intersect at a point in the direction of C and D.
Thus, AB is parallel to CD.
Q.E.D.
This proposition will be used in the proof of Proposition 28 which states that a line that intersects two parallel lines produces alternate angles that are equal to one another.