Sum of Two Sides of Triangle Greater than Third Side

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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

Triangle Inequality.png

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 theorem is Proposition $20$ of Book $\text{I}$ of Euclid's The Elements.
It is a geometric interpretation of the triangle inequality.


Sources