Relative Sizes of Sides of Obtuse Triangle

Theorem

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

Proof

 * Euclid-II-12.png

Let $\triangle ABC$ be an obtuse triangle where $\angle BAC$ is obtuse.

Produce $CA$ past $A$.

Construct a perpendicular $DB$ from $CA$ produced.

Then the square on $BC$ is greater than those on $BA$ and $AC$ by twice the rectangle contained by $CA$ and $AD$.

The proof is as follows.

From Square of Sum, the square on $DC$ equals the squares on $CA$ and $AD$ plus twice the rectangle contained by $CA$ and $AD$.

Add the square on $DB$ to each.

So the squares on $CD$ and $DB$ together equal the squares on $DB$, $CA$ and $AD$ plus twice the rectangle contained by $CA$ and $AD$.

But by Pythagoras's Theorem, the square on $CB$ equals the squares on $CD$ and $DB$ because $\angle BDC$ is a right angle.

Also by Pythagoras's Theorem, the square on $AB$ equals the squares on $AD$ and $DB$ because $\angle BDA$ is a right angle.

So the square on $CB$ equals the squares on $CA$ and $AB$ plus twice the rectangle contained by $CA$ and $AD$.

Hence the result.