Corresponding Angles Property

Theorem: Two lines are parallel if and only if the two angles of any pair of corresponding angles of any transversal are congruent (equal in measure).

Prerequisites:
Parallel lines (definition)
Angle on a straight line (proof)
Transitivity of parallelism (proof)

Proof:

First let us prove parallelism from the equality of corresponding angles:

Let AB and CD be two lines such that any pair of corresponding angles made by any transversal to them are congruent. Let EF be a transversal intersecting AB and CD at M and N respectively, such that $\angle EMB = 90^o$.

Since, corresponding angles are equal,

$\therefore\quad\;\;\angle END = \angle EMB = 90^o$

Also, since angle on a straight line is $180^o$, hence all the angles marked in the figure are $90^o$. So the figure is completely laterally symmetric with respect to the line EF, i.e., region to the left of the line EF is symmetric to the region to the right of the line EF.

Now, if possible, let us assume that the lines AB and CD are not parallel. Then by definition of parallel line, the two lines must intersect. Let us assume that the lines intersect somewhere on the left side of the line EF. Then, by the lateral symmetry about the line EF, they must also intersect somewhere on the right side of the line EF. But since the two straight lines can intersect each other at not more than one point, hence this is a contradiction.

Thus, the two line AB and CD are parallel.
Hence, equality of corresponding angles $\Rightarrow$ parallelism.

Conversely, let us assume that the lines AB and CD are parallel. We now show that any pair of corresponding angles made by any transversal to them are equal.

In the figure, let us assume that the line AB$\parallel$CD. Let EF be the transversal to them such that $\angle EMB = \alpha$ and $\angle END = \beta$. If possible, let us assume that $\alpha\neq\beta$. Then $\alpha>\beta$ or $\alpha<\beta$. Without any loss of generality, let us assume that $\alpha<\beta$. Then $\beta = \alpha + \gamma$ (say). Now let us draw a line A'B' such that the angle $\angle EMB' = \beta = \angle END$. Then by the result of the previous part,

$\qquad\quad A'B'\parallel CD$

Since AB$\parallel$CD, hence by the transitivity of parallelism property,

$\qquad\quad A'B'\parallel AB$

But this is a contradiction, as AB intersects A'B' at M.

Hence, $\alpha\not<\beta$.
Similarly, $\alpha\not>\beta$.

Hence, $\alpha = \beta$.

Therefore, parallelism $\Rightarrow$ equality of corresponding angles.

Hence, the theorem.

Recommended:
Alternate angles property
Vertical angle theorem
Interior angles property