# External Angle of Triangle equals Sum of other Internal Angles/Proof 1

## Contents

## Theorem

The external angle of a triangle equals the sum of the other two internal angles.

In the words of Euclid:

*In any triangle, if one of the sides be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of the triangle are equal to two right angles.*

(*The Elements*: Book $\text{I}$: Proposition $32$)

## Proof

Let $\triangle ABC$ be a triangle.

Let $BC$ be extended to a point $D$.

Construct $CE$ through the point $C$ parallel to the straight line $AB$.

We have that $AB \parallel CE$ and $AC$ is a transversal that cuts them.

From Parallelism implies Equal Alternate Angles:

- $\angle BAC = \angle ACE$

Similarly, we have that $AB \parallel CE$ and $BD$ is a transversal that cuts them.

From Parallelism implies Equal Corresponding Angles:

- $\angle ECD = \angle ABC$

Thus by Euclid's Second Common Notion:

- $\angle ACD = \angle ABC + \angle BAC$

$\blacksquare$

## Historical Note

This theorem is the first part of Proposition $32$ of Book $\text{I}$ of Euclid's *The Elements*.

Euclid's proposition $32$ consists of two parts: the first of which is this, and the second part of which is Sum of Angles of Triangle equals Two Right Angles.

It is not known for certain, but this theorem is traditionally ascribed to Pythagoras.

However, it is plausible that Pythagoras himself received it (either directly or indirectly) from Thales of Miletus, who may well have used it in his proof of what is now known as Thales' Theorem.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 1*(2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions