A second approach to a proof of the External Angle Theorem for triangles

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Hello, I have been given the task shown to solve. I have done so. I am now trying to find a second way of solving thus being able to anticipate any answer someone may have when solving.

I was able to prove this task 1 one by using the definition of a triangle and definition of supplementary angles and then substitution, i,e:

$\angle A + \angle B + \angle 2= 180^{\circ}$ by def of a triangle.

$\angle 2 + \angle 1 = 180^{\circ}$ by def of supplementary angles.

$\angle 2 =180^{\circ} - \angle 1$ by algebra

$\angle A + \angle B + (180^{\circ} - \angle 1)= 180^{\circ}$ by substitution

$\angle A + \angle B= \angle 1$

I am struggling of thinking of another approach that doesn't include a similar approach using supplementary angles. OR maybe the new approach does use that concept, but then I'm not sure to use it differently than the first. Any help?

(I understand that some will say this is homework, but rather I am trying to anticipate all answers that could be produced for this task).

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Draw a ray CD parallel to AB cutting angle 1 into two pieces, then use corresponding/alternate angles of parallel lines by a transversal line.