# 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.

In the words of Euclid:

In any triangle two sides taken together in any manner are greater than the remaining one.

## Proof Let $ABC$ be a triangle.

By Euclid's Second Postulate, we can produce $BA$ past $A$ in a straight line.

By Construction of Equal Straight Lines from Unequal, there exists a point $D$ such that $DA = CA$.

Therefore, from Isosceles Triangle has Two Equal Angles:

$\angle ADC = \angle ACD$

Thus by Common Notion $5$:

$\angle BCD > \angle BDC$

Thus $\triangle DCB$ is a triangle having $\angle BCD$ greater than $\angle BDC$

$BD > BC$

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 proof is Proposition $20$ of Book $\text{I}$ of Euclid's The Elements.
It is a geometric interpretation of the Triangle Inequality.