Square equals Sum of Squares implies Right Triangle

Theorem
If, in a triangle, the square on one of the sides equals the sum of the squares on the other two sides, the angle adjacent to those other two sides is a right angle.

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.