(by Polymath) This page is part of the Advanced High School Math Project.

This proof of the Pythagorean theorem uses the power theorem for circles, which is not as well-known as it should be. It says this:

Given a circle and a point not on it (interior or exterior), draw a line that goes through the point and intersects the circle twice. Those two intersections are then each paired with the given point to form two segments (that fall on the line). The product of the lengths of those segments is a constant for that circle and point; which is to say, the product is independent of the choice of line. If one (or, trivially, both) of the lines are tangent to the circle, you can use the one segment instead, and the square of that length equals the same constant. That constant is called the power of the point.

Crucially, the proof of the power theorem uses similar triangles (it's not actually that hard), so that means that using it to prove the Pythagorean theorem doesn't constitute circular logic. I do not claim to be the first to find this proof, of course, but I did discover it on my own. I have since seen it again in only one other place on the internet, and then only near the end of a LONG list of Pythagorean theorem proofs. But it's very short and efficient, and uses the power theorem a little unexpectedly, so I thought it would be instructive.

Given the right triangle ABC, labeled in the standard manner, draw a circle centered at B with a radius of *a*.
Then extend the hypotenuse to the edge of the circle (point X). Since
angle C is right, AC is a tangent to the circle, and we can use the
power theorem for point A. We will square *b* since it is the only distance to the circle. And since the radius of the circle is *a*, the two required distances along AX from A to each of the intersections with the circle) are *c-a* and *c+a*. The power theorem says that the square of *b* equals the product of those two lengths. This gives:

which is exactly what we wanted.

Very efficient.