Relative Sizes of Sides of Acute Triangle

Theorem

In the words of Euclid:

In acute-angled triangles the square on the side subtending the acute angle is less than the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle, namely that on which the perpendicular falls, and the straight line cut off within by the perpendicular towards the acute angle.

(The Elements: Book $\text{II}$: Proposition $13$)

Proof

Let $\triangle ABC$ be an acute triangle, and so by definition, $\angle ABC$ is acute.

Construct a perpendicular $AD$ from $BC$.

Then the square on $AC$ is less than those on $AB$ and $BC$ by twice the rectangle contained by $BC$ and $BD$.

The proof is as follows.

From Square of Difference, squares on $BC$ and $BD$ equal the square on $DC$ plus twice the rectangle contained by $BC$ and $BD$.

Add the square on $AD$ to each.

So the squares on $BC$, $BD$ and $AD$ together equal the squares on $DC$ and $AD$ plus twice the rectangle contained by $BC$ and $BD$.

But by Pythagoras's Theorem, the square on $AB$ equals the squares on $AD$ and $BD$ because $\angle ADB$ is a right angle.

Also by Pythagoras's Theorem, the square on $AC$ equals the squares on $AD$ and $DC$ because $\angle ADC$ is a right angle.

So the squares on $AB$ and $BC$ equals the square on $AC$ plus twice the rectangle contained by $BC$ and $BD$.

Hence the result.

$\blacksquare$

Historical Note

This proof is Proposition $13$ of Book $\text{II}$ of Euclid's The Elements.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 1 (2nd ed.) ... (previous) ... (next): Book $\text{II}$. Propositions