### Theory:

Axiom $$3$$: If the transversal intersects two parallel lines, then each of the corresponding angles are equal.

This axiom is also known as the corresponding angles axiom.

Let us look at the converse of this axiom.

If a transversal line intersects any two lines and the corresponding lines are equal, then the two lines are parallel.

Let us try to verify the converse of axiom $$3$$.

Let us draw a line $$PQ$$ and have the points $$M$$ and $$N$$ on it. Through $$M$$ and $$N$$, let us draw two rays $$MA$$ and $$NC$$ respectively, as shown in the figure.

Then let us extend the rays $$MA$$ and $$NC$$ to the other side of $$PQ$$, such that the lines $$AB$$ and $$CD$$ are availed.

We can also draw two common perpendicular lines to $$AB$$ and $$CD$$ and measure the lengths. The lengths will be equal everywhere.

Thus, the converse of axiom $$3$$ is verified, and it also leads to the following axiom.

Axiom $$4$$: If a transversal intersects two lines such that a pair of corresponding angles are equal, then the two lines are parallel to each other.

Let us draw two lines $$AB$$ and $$CD$$, and a transversal $$PQ$$. Let the points at which $$PQ$$ meets $$AB$$ and $$CD$$ be $$M$$ and $$N$$ respectively, as shown in the figure given below.

We should prove that the lines AB and CD are parallel are equal.

Hence, we need to prove that the alternate interior angles are equal.

Thus, we should prove that $$\angle AMN = \angle DNM$$, and $$\angle BMN = \angle CNM$$.

We know that, $$\angle PMA = \angle MNC \longrightarrow (1)$$

[By corresponding angles axiom]

$$\angle PMA = \angle BMN \longrightarrow (2)$$

[By vertically opposite angles theorem]

On equating $$(1)$$ and $$(2)$$, we get:

$$\angle MNC = \angle BMN$$

Similarly, $$\angle AMN = \angle MND$$

This result of axiom $$4$$ is stated in theorem $$2$$, and the theorem is discussed in the next section.