The exterior angle is greater than the non-adjacent interior angles

The first version of the exterior angle theorem is an inequality, and it holds completely generally, without needing to know the sum of the interior angles.

Theorem — Exterior angle (inequality)

In every triangle, an exterior angle is greater than each of the two non-adjacent interior angles.

Proof

The construction: $M$ the midpoint of $BC$, $MD\cong AM$; the angle $\widehat{DBM}$ is only a part of the exterior angle at $B$.

Success

1. Construction. Let $M$ be the midpoint of $BC$. We join $A$ to $M$ and extend beyond $M$ by a segment $MD\cong AM$. We join $D$ to $B$.
2. I consider. The triangles $ACM$ and $DBM$: $AM\cong MD$ (construction), $CM\cong MB$ ($M$ midpoint), $\widehat{AMC}\cong\widehat{DMB}$ (vertically opposite).
3. I deduce. By the first criterion: $\triangle ACM\cong\triangle DBM$, therefore $\widehat{ACM}\cong\widehat{DBM}$.
4. I observe. Now, $\widehat{DBM}$ is only a part of the exterior angle at $B$, therefore the exterior angle is greater than $\widehat{ACM} = \widehat{ACB}$.

$\blacksquare$

Links

Topics: Euclidean geometry
Concepts: Vertically opposite angles · Exterior angle · Congruence criteria · Proof
Skills: Proving · Synthetic geometry