Sum of Angles of Triangle equals Two Right Angles

Theorem

In a triangle, the sum of the three interior angles equals two right angles.

In the words of Euclid:

In any triangle, if one of the sides be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of the triangle are equal to two right angles.

Proof 1

Let $\triangle ABC$ be a triangle.

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

$\angle ACD = \angle ABC + \angle BAC$
$\angle ACB + \angle ACD = \angle ABC + \angle BAC + \angle ACB$

But from Two Angles on Straight Line make Two Right Angles, $\angle ACB + \angle ACD$ equals two right angles.

So by Euclid's First Common Notion, $\angle ABC + \angle BAC + \angle ACB$ equals two right angles.

$\blacksquare$

Proof 2

Let $\Delta ABC$ be a triangle.

Let $DAE$ be a line such that $DE \parallel BC$.

$\angle DAB = \angle ABC$

and:

$\angle EAC = \angle ACB$

Therefore, the sum of the three angles is:

$\angle ABC + \angle BCA + \angle CAB = \angle DAB + \angle BAC + \angle CAE = 180 \degrees$

$\blacksquare$

Historical Note

This proof is the second part of Proposition $32$ of Book $\text{I}$ of Euclid's The Elements.
Euclid's proposition $32$ consists of two parts, the first of which is External Angle of Triangle equals Sum of other Internal Angles, and the second part of which is this.