Everyone knows that A² + B² = C², but can you prove it? There are at least 88 ways to do it. Here's my personal favorite.

The Pythagorean Theorem is known by anyone who has taken basic geometry. In a right triangle, the square of the hypotenuse is equal to the sum of the squares of both legs. It's so famous that it showed up in *The Wizard of Oz* movie (albeit slightly misquoted). But how to prove it?

It's relatively easy to show that the equation holds up through grunt work, making triangles and measuring them out. Still, that will only prove that the Pythagorean Theorem holds up in most cases, not in every case, no matter what.

Luckily Cut-the-knot.org has not one, not two but eighty ways that the Pythagorean Theorem has been proved over the years. You could spend almost three months learning a proof a day, but this is my favorite.

This was done with by a high school kid in the 1930s, and is a really gorgeous idea.

Most people know that the area of a rectangle is determined by multiplying base times height. The area of the blue square, then, is the hypotenuse of one triangle multiplied by the hypotenuse of the other. Since all the triangles are the same, the area of the blue square is the length of the hypotenuse squared.

The square on the right shifts the triangles around, leaving two red squares visible. The area of the small upper red square is defined on all sides by the short side of the triangles, so it can be expressed as length of the short side squared. The area of larger, lower red square is defined on all sides by the longer side of the triangles, so it can be expressed as the length of the long side squared.

Shifting things around doesn't give them a larger or smaller area. In other words, the non-yellow area of both squares is the same, or equal.

(short side)² + (long side)² = (hypotenuse)²

Below is a slightly different view of the shifting pieces, which gives the same result.

If you don't like it, go to the site and pick another proof. There are plenty to choose from.

Via cut-the-knot.org.