Sum of the interior angles of a triangle

Here is the first great application of alternate interior angles: the elegant and very short proof that the sum of the interior angles of a triangle is always a straight angle.

Theorem — Sum of the interior angles of a triangle

In every triangle the sum of the interior angles is $\pi$ (that is, $180^\circ$). $\alpha,\,\beta,\,\gamma \text{ angoli interni di } \triangle ABC \implies \alpha + \beta + \gamma = \pi$

Proof

Through $C$ we draw the parallel $DE$ to $AB$: the base angles “climb back” around $C$ forming a straight angle.

Success

1. Construction. Through $C$ we draw the line $DE$ parallel to $AB$.
2. Observe. Since $AB \parallel DE$ and $AC$ is a transversal, the angles $\widehat{BAC}$ and $\widehat{ACD}$ are alternate interior, hence $\widehat{ACD} \cong \alpha$.
3. Similarly. Using $BC$ as the transversal: $\widehat{ECB} \cong \beta$.
4. Deduce. At this point, the angles $\widehat{ACD}$, $\widehat{ACB} = \gamma$ and $\widehat{ECB}$ are consecutive and form a straight angle: $\alpha + \gamma + \beta = \pi.$

$\blacksquare$

Links

Topics: Euclidean geometry
Concepts: Alternate interior angles · Proof · Line · Parallel lines · Sum of triangle angles
Skills: Proving · Synthetic geometry