# Two Straight Lines make Equal Opposite Angles

## Theorem

In the words of Euclid:

*If two straight lines cut one another, they make the vertical angles equal to one another.*

(*The Elements*: Book $\text{I}$: Proposition $15$)

### Porism

It follows that if two straight lines cut one another, the angles at the point of intersection make four right angles.

## Proof

Let $AB$ and $CD$ be two straight lines that cut each other at the point $E$.

Since the straight line $AE$ stands on the straight line $CD$, the angles $\angle AED$ and $\angle AEC$ make two right angles.

Since the straight line $DE$ stands on the straight line $AB$, the angles $\angle AED$ and $\angle BED$ make two right angles.

But $\angle AED$ and $\angle AEC$ also make two right angles.

So by Common Notion 1 and the fact that all right angles are congruent, $\angle AED + \angle AEC = \angle AED + \angle BED$.

Let $\angle AED$ be subtracted from each.

Then by Common Notion 3 it follows that $\angle AEC = \angle BED$.

Similarly it can be shown that $\angle BEC = \angle AED$.

$\blacksquare$

## Also known as

This result is also called the **vertical angle theorem**.

The arises from the fact that the angles proven equal are known as vertical angles.

## Historical Note

This proof is Proposition $15$ of Book $\text{I}$ of Euclid's *The Elements*.

It appears to have originally been created by Thales of Miletus.

## 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 $1.3$ - 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.1$: Thales (ca. $\text {625}$ – $\text {547}$ B.C.)