A few years ago, my dad asked me a question related to his job (for a mobile phone company at the time) that he thought i might be able to attack with my geometry skills. He's pretty good at math himself (Ph.D. in physics), but his knowledge of the basics isn't very up-to-date. Anyway, it turned out that I couldn't help him much, but in the course of trying, I used the power theorem for circles, which he was unfamiliar with. 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; 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.

Anyway, my dad had never heard of this theorem, and took some time to try to prove it himself--the proof uses similar triangles, and isn't all that hard, but my father's geometry is rusty, and he couldn't prove it. But in his notes he had a picture that I tried to analyze, and in that process I discovered a proof of the Pythagorean theorem that I had never seen before. I do not claim to be the first to find it, 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 I'd put it down here.

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 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.