Square equals Sum of Squares implies Right Triangle

Proof

 * Euclid-I-48.png

Let $\triangle ABC$ be a triangle such that the square on $BC$ equals the sum of the squares on $AB$ and $AC$.

Construct $AD$ perpendicular to $AC$.

Make $AD$ equal to $AB$, and join $CD$.

Since $DA = AB$, the square on $AD$ equals the square on $AB$.

Add the square on $AC$ to each.

Then the squares on $AD$ and $AC$ equal the squares on $AB$ and $AC$.

But $\angle DAC$ is a right angle.

Therefore, by Pythagoras's Theorem, the square on $DC$ equals the sum of the squares on $AD$ and $AC$.

Also, the square on $BC$ equals the sum of the squares on $AB$ and $AC$, by hypothesis.

So the square on $DC$ equals the square on $BC$, and so $DC = BC$.

So from Triangle Side-Side-Side Equality, $\triangle ABC = \triangle DAC$.

But as $\triangle DAC$ is a right triangle, then so must $\triangle ABC$ be.