Exterior angle theorem

Statement

Let $ABC$ be a triangle in a Euclidean space $\mathbb{R}^n$, and let $\angle$ denote the angle at a point. Let $D$ be a point such that $C$ lies strictly between $B$ and $D$ on the line $BC$. Then the exterior angle at $C$ is the sum of the two interior angles at the other vertices: $$\angle ACD = \angle BAC + \angle CBA.$$

Remarks

The angles $\angle ACD$ and $\angle ACB$ add up to $\pi$. Thus the statement follows from the angle sum of a triangle. A consequence is that the exterior angle is greater than each of the two remote interior angles. This weaker form does not need the parallel postulate.

History

In Book I of Euclid's Elements, the weaker form is Proposition 16. The form with the sum is part of Proposition 32 of the same book. Many school books call the form with the sum the exterior angle theorem.

Explanations

L1 human

Extend one side of a triangle past a corner. The angle outside the triangle at that corner is the exterior angle. It is exactly as large as the two far angles of the triangle together.