Is being a right triangle both necessary and sufficient for the Pythagorean Theorem to hold?

Solution 1:

Suppose that $a^2 + b^2 = c^2$ is a necessary condition to be a right triangle.

Corollary: $a^2 + b^2 = c^2$ is a sufficient condition to be a right triangle.

Proof: Suppose you're given a particular triangle whose side lengths satisfy $a^2 + b^2 = c^2$.

Construct a right triangle whose two sides adjacent to the angle are of length $a$ and $b$ (e.g. start with the right angle, and mark off the two sides to have the appropriate length). By the supposition, it follows that the third side of the right triangle has length $c$.

By the side-side-side congruence theorem, the original triangle is congruent to the right triangle, and thus the original triangle is a right triangle.

Solution 2:

Yep, the converse of the Pythagorean Theorem is also true. It can be proved using the law of cosines:

Assume $a,b,c$ are the sides of a triangle and they satisfy $a^2 + b^2 = c^2$. Let $\angle ACB = \gamma$. By the law of cosines,

$$ c^2 = a^2 + b^2 - 2ab\cos\gamma . $$

Since $a,b > 0$, we must have $\cos\gamma = 0$. Since $0 < \gamma < 180^\circ$, we must have $\gamma = 90^\circ$. So $\Delta ABC$ is a right triangle.

Solution 3:

The Pythagorean theorem gives a necessary and sufficient characterization of being a right triangle. In other words, when $a$, $b$, and $c$ are the side lengths of a triangle, $a^2 + b^2 = c^2$ is equivalent to the angle opposite the $c$-side being a right angle. This follows from the law of cosines, which states

$$ c^2 = a^2 + b^2 - 2 ab \cos C $$

where $C$ is the angle opposite the $c$-side, for any triangle with side lengths $a$, $b$, and $c$. We only get $c^2 = a^2 + b^2$ when $\cos C = 0$, or when $C = 90^{\circ}$.

It is also interesting to note that this theorem (both the necessary and sufficient parts) are proved in Euclid's Elements via geometric methods.