Relative Sizes of Sides of Acute Triangle

Proof

 * Euclid-II-13.png

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.