Sum of Internal Angles of Triangle

Theorem
The sum of the three internal angles of a triangle is $180^\circ$.

Proof


Let $\Delta ABC$ be a triangle.

Let $DAE$ be a line such that $DE//BC$.

By parallelism implies equal corresponding angles, $\angle DAB = \angle ABC$ and $\angle EAC = \angle ACB$.

Therefore, the sum of the three angles is $\angle ABC + \angle BCA + \angle CAB = \angle DAB + \angle BAC + \angle CAE = 180^\circ$.