Consider a triangle with A, B, and C, and one of the exterior angles corresponding to ∠C.
The diagram shows that
∠C and
∠PCA form a , so the sum of their measures is
180∘. Additionally, by the , the sum of the angle measures of
△ABC is
180∘.
{m∠C+m∠PCA=180∘m∠A+m∠B+m∠C=180∘(I)(II)
Now
m∠C can be isolated in Equation (I).
m∠C+m∠PCA=180∘⇕m∠C=180∘−m∠PCA
Next, the expression of
m∠C can be substituted into Equation (II).
m∠A+m∠B+m∠C=180∘
m∠A+m∠B+(180∘−m∠PCA)=180∘
m∠A+m∠B+180∘−m∠PCA=180∘
m∠A+m∠B−m∠PCA=0
m∠A+m∠B=m∠PCA
m∠PCA=m∠A+m∠B
It has been proven that the measure of
∠PCA is equal to the sum of the measures of
∠A and
∠B. Therefore, it can be said that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent interior angles.