Relative Sizes of Sides of Acute Triangle

Theorem

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

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.