Equal Alternate Angles implies Parallel Lines

Theorem

Given two infinite straight lines which are cut by a transversal, if the alternate angles are equal, then the lines are parallel.

In the words of Euclid:

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.

(The Elements: Book $\text{I}$: Proposition $27$)

Proof

Let $AB$ and $CD$ be two infinite straight lines.

Let $EF$ be a transversal that cuts them.

Let at least one pair of alternate angles be equal.

Without loss of generality, let $\angle AHJ = \angle HJD$.

Aiming for a contradiction, suppose that $AB$ and $CD$ are not parallel.

Then they meet at some point $G$.

Without loss of generality, let $G$ be on the same side as $B$ and $D$.

Since $\angle AHJ$ is an exterior angle of $\triangle GJH$, from External Angle of Triangle is Greater than Internal Opposite, $\angle AHJ > \angle HJG$, which is a contradiction.

Similarly, they cannot meet on the side of $A$ and $C$.

Therefore, by definition, $AB$ and $CD$ are parallel.

$\blacksquare$

Historical Note

This proof is Proposition $27$ of Book $\text{I}$ of Euclid's The Elements.
It is the converse of the first part of Proposition $29$: Parallelism implies Equal Alternate Angles.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 1 (2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions
• 1968: M.N. Aref and William Wernick: Problems & Solutions in Euclidean Geometry ... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.4$