Book 1 – Proposition 20

In any triangle two sides taken together in any manner are greater than the remaining one.
[Desmos graphs available here]

Intuitively this should make sense, because if the two sides were shorter than the third they would not even cover the third segment laid end to end, making it impossible to form a triangle.

For a triangle ABC, we will prove the sides AB and AC are greater than side BC. A similar strategy can be used to show that AB and BC are greater than AC, and that AC and BC are greater than AB.

Extend the segment BA through a point D such that the segment AD has the same length as segment AC. (Proposition 2 allows us to construct a line segment of the same length as AC.)
Let DC be joined.

Look at triangle ACD. Since AD is equal to AC, triangle ACD is an isosceles triangle. By Proposition 5, angle ADC is equal to angle ACD. (The base angles of an isosceles triangle are equal.)

Because angle BCD is greater than angle ACD, we know that angle BCD is greater than angle ADC. We also know that angles ADC and BDC are equal.

Looking at triangle BCD …
Since angle BCD is greater than angle BDC, side BD is greater than side BC by Proposition 19 (the greater angle is subtended by the greater side).
The side BD is equal to AD and AB.
Since AD was created equal to AC, side BD is equal to AC and AB.
Thus AC and AB together are greater than BC.
[ BD > BC … AD & AB together > BC … AC & AB > BC]

Q.E.D.

We showed that two arbitrarily selected sides together are greater than the remaining side, so our results apply to each pairing in the triangle.

Questions? Comments? You know what to do – George