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 $$BA + BC > AC$$.

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

It is also a geometric interpretation of the triangle inequality.