Sum of Two Sides of Triangle Greater than Third Side

Theorem

Given a triangle $ABC$, the sum of the lengths of any two sides of the triangle is greater than the length of the third side.

Proof

We can extend $BA$ past $A$ into a straight line.

There exists a point $D$ such that $DA = CA$.

Therefore, $\angle ADC = \angle ACD$ because isosceles triangle have two equal angles.

Thus, $\angle BCD > \angle BDC$ by Euclid's fifth common notion.

Since $\triangle DCB$ is a triangle having $\angle BCD$ greater than $\angle BDC$, this means that $BD > BC$.

But $BD = BA + AD$, and $AD = AC$.

Thus, $BA + AC > BC$.

A similar argument shows that $AC + BC > BA$ and $BA + BC > AC$.

$\blacksquare$

Historical Note

This is Proposition 20 of Book I of Euclid's The Elements.

It is a geometric interpretation of the triangle inequality.