Relative Sizes of Sides of Obtuse Triangle

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.