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.
[Desmos graphs can be found here]
Start with triangle ABC, and let side BC be produced to D.
We will prove that (i) the exterior angle ACD is equal to the two interior and opposite angles ABC and CAB, and (ii) the three interior angles ABC, BCA, and CAB are equal to two right angles.
Using Proposition 31 (which covered how to construct a line through a given point that is parallel to a given line), let CE be drawn through point C that is parallel to AB.
Line AC falls upon the two parallel lines AB and CE, so by Proposition 29 (a straight line falling on two parallel lines makes the alternate angles equal to one another) the alternate angles BAC and ACE are equal to one another.
Line BD also falls upon the two parallel lines AB and CE, so Proposition 29 (exterior angle is equal to interior and opposite angle) tells us that the exterior angle ECD is equal to the opposite and interior angle ABC.
Angle ACD is made up of the angles ACE (which is equal to angle BAC) and ECD (which is equal to angle ABC). So, the exterior angle ACD is equal to the two interior and opposite angles ABC and BAC.
Moving on to the proof of (ii), add the angle ACB to both angle ACD as well as to the angles ABC and BAC.
Angles ACD and ACB are equal to the angles ABC, BAC, and ACB.
By Proposition 13 (a straight line set up on a straight line makes two angles that are equal to two right angles) angles ACD and BAC are equal to two right angles.
Therefore the three interior angles ABC, BAC, and ACB are equal to two right angles.
Q.E.D.
That’s two-thirds of the way through Book 1. Comments or questions? I’d love to hear them – George