# Two Angles making Two Right Angles make Straight Line

## Contents

## Theorem

In the words of Euclid:

*If with any straight line, and at a point on it, two straight lines not lying on the same side make the adjacent angles equal to two right angles, the two straight lines will be in a straight line with one another.*

(*The Elements*: Book $\text{I}$: Proposition $14$)

## Proof

Let $AB$ be the given straight line and $B$ the given point on it.

Let $BC$ and $BD$ be the two straight lines not lying on the same side of it such that $\angle ABC + \angle ABD$ equal two right angles.

Suppose $BD$ is not in a straight line with $BC$.

Now some line must be, so let $BE$ be in a straight line with $BC$ instead.

Then since $AB$ stands on the (supposed) straight line $CBE$, $\angle ABC + \angle ABE$ equal two right angles.

But we already know that $\angle ABC + \angle ABD$ equal two right angles.

Therefore, by Common Notion 1 and the fact that all right angles are congruent, $\angle ABC + \angle ABD = \angle ABC + \angle ABE$.

So let $\angle ABC$ be subtracted from each.

Then by Common Notion 3, the remaining angle $\angle ABE$ equals remaining angle $\angle ABD$.

But $\angle ABD$ is greater than $\angle ABE$, so this is impossible.

So $BE$ can not be in a straight line with $BC$.

Similarly we can show that *any* straight line which is not $BD$ can not be in a straight line with $BC$.

Therefore $BC$ is in a straight line with $BD$, hence the result.

$\blacksquare$

## Historical Note

This theorem is Proposition $14$ of Book $\text{I}$ of Euclid's *The Elements*.

It is the converse of Proposition $13$: Two Angles on Straight Line make Two Right 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 $1.2$