Book 1 – Proposition 16

In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior or opposite angles.
[Desmos graphs can be found here]

Start with triangle ABC, and let one side of BC be produced to point D.

We will prove that the exterior angle ACD is greater than either of the interior and opposite angles BAC and CBA.

By Proposition 10 (that allows us to bisect a line segment) let AC be bisected at E.
Let BE be joined and produced to a point F so that EF is equal to BE. (Proposition 3 allows us to construct a line segment equal to a given segment.)
Let FC be joined.

Finally, let the side AC be drawn through to a point G as shown.

We will now examine the two triangles ABE and CFE.

• Sides AE and EC are equal, since E bisects AC.
• Sides BE and EF are equal according to the way we selected point F.
• Angles AEB and CEF are equal to each other because they are vertical angles.
  (Proposition 15 states that vertical angles are equal.)

So, by Proposition 4 (“side-angle-side”), the side AB is equal to the side CF and the remaining corresponding angles are equal. In particular, angle BAE is equal to angle ECF.

Angle ECD is greater than angle ECF, and that implies that angle ACD is greater than angle BAE. (Angle ACD is equal to angle ECD, and angle BAE is equal to angle ECF.)

In a similar fashion, we can prove that angle ACD is greater than angle ABC.
(Bisect BC … angle BCG equals angle ACD (vertical angles) … construct triangles and show that angle BCG is greater than angle ABC.)

Q.E.D.

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Later we will see that the three interior angles ABC, BCA, and ACB are equal to two right angles.
The adjacent angles ACB and ACD are also equal to two right angles.
That shows that the two angles ABC and BCA are equal to the angle ACD, and angle ACD must be greater than either of those two angles.

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If you have feedback, I’d love to hear it – George