# External Angle of Triangle Greater than Internal Opposite

## Theorem

The external angle of a triangle is greater than either of the opposite internal angles.

In the words of Euclid:

*In any triangle, if one of the sides be produced, the The exterior angle is greater than either interior and opposite angles.*

(*The Elements*: Book $\text{I}$: Proposition $16$)

## Proof

Let $\triangle ABC$ be a triangle.

Let the side $BC$ be extended to $D$.

Let $AC$ be bisected at $E$.

Let $BE$ be joined and extended to $F$.

Let $EF$ be made equal to $BE$.

(Technically we really need to extend $BE$ to a point beyond $F$ and then crimp off a length $EF$.)

Let $CF$ be joined.

Let $AC$ be extended to $G$.

We have $\angle AEB = \angle CEF$ from Two Straight Lines make Equal Opposite Angles.

Since $AE = EC$ and $BE = EF$, from Triangle Side-Angle-Side Equality we have $\triangle ABE = \triangle CFE$.

Thus $AB = CF$ and $\angle BAE = \angle ECF$.

But $\angle ECD$ is greater than $\angle ECF$.

Therefore $\angle ACD$ is greater than $\angle BAE$.

Similarly, if $BC$ were bisected, $\angle BCG$, which is equal to $\angle ACD$ by Two Straight Lines make Equal Opposite Angles, would be shown to be greater than $\angle ABC$ as well.

Hence the result.

$\blacksquare$

## Historical Note

This theorem is Proposition $16$ of Book $\text{I}$ of Euclid's *The Elements*.

## 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 $1.8$: Corollary $2$