# 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:

(*The Elements*: Book $\text{I}$: Proposition $20$)

## Proof

Let $ABC$ be a triangle

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

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

Therefore, from Isosceles Triangle has Two Equal Angles:

- $\angle ADC = \angle ACD$

Thus by Euclid's fifth common notion:

- $\angle BCD > \angle BDC$

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 proof is Proposition $20$ of Book $\text{I}$ of Euclid's *The Elements*.

It is a geometric interpretation of the triangle inequality.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 1*(2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions - 1968: M.N. Aref and William Wernick:
*Problems & Solutions in Euclidean Geometry*... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.18$ - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next): Entry:**triangle inequality**(for points in the plane)