External Angle of Triangle Greater than Internal Opposite

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

Proof

 * Euclid-I-16.png

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.