External Angle of Triangle equals Sum of other Internal Angles/Proof 2

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The external angle of a triangle equals the sum of the other two internal 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.

(The Elements: Book $\text{I}$: Proposition $32$)


Let $\triangle ABC$ be a triangle.

From Sum of Angles of Triangle equals Two Right Angles: Proof 2:

$\paren 1: \angle ABC + \angle BCA + \angle CAB = 180^\circ$

Extend $AB$ to $D$.

By Two Angles on Straight Line make Two Right Angles:

$\paren 1: \angle ABC + \angle CBD = 180^\circ$

Combining $\paren 1$ and $\paren 2$ and using Equality is Transitive:

$\angle ABC + \angle BCA + \angle CAB = \angle ABC + \angle CBD$

By using commom notion 3:

$\angle BCA + \angle CAB = \angle CBD$

By using Equality is Symmetric:

$\angle CBD = \angle BCA + \angle CAB$