# External Angle of Triangle equals Sum of other Internal Angles

## 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 1

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:

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

$\blacksquare$

## Proof 2

Let $\triangle ABC$ be a triangle.

From Sum of Angles of Triangle equals Two Right Angles: Proof 2:

- $\paren 1: \angle ABC + \angle BCA + \angle CAB = 180^\circ$

Extend $AB$ to $D$.

By Two Angles on Straight Line make Two Right Angles:

- $\paren 1: \angle ABC + \angle CBD = 180^\circ$

Combining $\paren 1$ and $\paren 2$ and using Equality is Transitive:

- $\angle ABC + \angle BCA + \angle CAB = \angle ABC + \angle CBD$

By using commom notion 3:

- $\angle BCA + \angle CAB = \angle CBD$

By using Equality is Symmetric:

- $\angle CBD = \angle BCA + \angle CAB$

$\blacksquare$

## Also known as

This theorem is referred to by some sources as the **Interior Angles Theorem**, and by others as the **Exterior Angle Theorem**. These names, however, are also used to refer to other theorems.

In the wake of such possible confusion, it is prudent not to refer to it by name at all, but by description, as has been done on $\mathsf{Pr} \infty \mathsf{fWiki}$.

## 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

- 1968: M.N. Aref and William Wernick:
*Problems & Solutions in Euclidean Geometry*... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.8$ - 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.1$: Thales (ca. $\text {625}$ – $\text {547}$ B.C.) - 1998: David Nelson:
*The Penguin Dictionary of Mathematics*(2nd ed.) ... (previous) ... (next):**triangle**: $(3)$ - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next):**triangle**: $(3)$