External Angle of Triangle is Greater than Internal Opposite

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

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.