### Theory:

Theorem $$2$$: If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.

Let us consider the figure given below.

Let us show that the pair of alternate angles are equal using the converse of corresponding angles theorem.

The transversal $$PQ$$ intersects the lines $$AB$$ and $$CD$$ at $$M$$ and $$N$$ respectively.

From the figure, it is also understood that $$\angle AMN$$ equals $$\angle MND$$.

We should prove that $$AB \parallel CD$$.

$$\angle AMN =\angle PMB \longrightarrow (1)$$

[By the vertically opposite angles theorem]

$$\angle AMN =\angle MND \longrightarrow (2)$$ [Given]

From $$(1)$$ and $$(2)$$, it is understood that $$\angle PMB = \angle MND$$.

The angles $$\angle PMB$$ and $$\angle MND$$ are corresponding angles.

Thus, $$AB \parallel CD$$.

[By the converse of corresponding angles axiom]

The result of Theorem $$2$$ leads to the statement of Theorem $$3$$.

Theorem $$3$$: If a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel.

In the statement, it is given that alternate interior angles are equal.

Here, the alternate interior angles are $$\angle AMN$$ and $$\angle MND$$.

That is, $$\angle AMN = \angle MND \longrightarrow (1)$$

By the vertically opposite angles theorem, 'When two lines intersect, then the vertically opposite angles are equal.'

Here, $$AB$$ and $$PQ$$ intersect each other. Also, one pair of vertically opposite angles are $$\angle AMN$$ and $$\angle PMB$$.

That is, $$\angle AMN = \angle PMB \longrightarrow (2)$$

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

$$\angle MND = \angle PMB \longrightarrow (3)$$

$$\angle MND$$ and $$\angle PMB$$ are corresponding angles. Such angles are equal only if two parallel lines intersect with a transversal.

Here, $$PQ$$ is a transversal.

Therefore, $$AB$$ and $$CD$$ are parallel to each other.

Theorem $$4$$: If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.

Let us look at the figure required to prove this theorem.

The transversal $$PQ$$ intersects the two parallel lines $$AB$$ and $$CD$$.

We should prove that:

1. $$\angle AMN + \angle MNC = 180^\circ$$

(Or)

2. $$\angle BMN + \angle MND = 180^\circ$$

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

[By corresponding angles axiom]

$$\angle PMA + \angle AMN = 180^\circ \longrightarrow (2)$$

[By linear pair axiom]

By substituting $$(1)$$ in $$(2)$$, we get:

$$\angle MNC + \angle AMN = 180^\circ$$

Similarly, we can prove that $$\angle BMN + \angle MND = 180^\circ$$.

Theorem $$5$$: If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel.

Let us look at the figure required to prove this theorem.

The transversal $$PQ$$ intersects the lines $$AB$$ and $$CD$$ at $$M$$ and $$N$$ respectively.

We should prove that, $$AB \parallel CD$$.

$$\angle AMN + \angle MNC = 180^\circ \longrightarrow (1)$$ [Given]

$$\angle PMA + \angle AMN = 180^\circ \longrightarrow (2)$$

[By linear pair axiom]

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

$$\angle PMA = \angle MNC$$

The angles $$\angle PMA$$ and $$\angle MNC$$ are corresponding angles.

Thus by the converse of corresponding angles axiom, we can prove that $$AB \parallel CD$$.