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.

(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 $1.18$
• 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Entry: triangle inequality: 1.
• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Entry: triangle inequality (for points in the plane)