External Angle of Triangle equals Sum of other Internal Angles/Proof 1

Proof

 * Euclid-I-32.png

Let $\triangle ABC$ be a triangle.

Let $BC$ be extended to a point $D$.

Construct $CE$ through the point $C$ parallel to the straight line $AB$.

We have that $AB \parallel CE$ and $AC$ is a transversal that cuts them.

From Parallelism implies Equal Alternate Angles:
 * $\angle BAC = \angle ACE$

Similarly, we have that $AB \parallel CE$ and $BD$ is a transversal that cuts them.

From Parallelism implies Equal Corresponding Angles:
 * $\angle ECD = \angle ABC$

Thus by Euclid's Second Common Notion:
 * $\angle ACD = \angle ABC + \angle BAC$