1

My favorite is Euclid's original proof for two reasons:

First, it requires minimal raw material. It only needs the result that the area of a triangle is half the area of a rectangle with the same base and altitude.

Second, it gives additional information, in that it shows how to divide the square on the hypotenuse into two rectangles each of which is equal in area to one of the squares on the legs.

So, what's yours?

marty cohen
  • 107,799
  • Take a triangle $ABC$ with right angle at $C$. Drop a perpendicular from $C$ to $P$ on $AB$, This divides the original triangle into two triangles similar to the original. The area of the original is $\kappa c^2$ for some constant $\kappa$. By a scaling argument the two little triangles have area $\kappa a^2$ and $\kappa b^2$, and it is clear that $\kappa c^2=\kappa a^2+\kappa b^2$. – André Nicolas Jul 13 '15 at 22:11

2 Answers2

2

Let $ABC$ be a right triangle, with right angle at $A$. Let $H$ be the foot of the altitude from $A$. Triangles $ABH$, $ACH$ and $ABC$ are similar, being their areas proportional to the square of the hypothenuses, that is, $$a^2=b^2+c^2$$

For brevity.

ajotatxe
  • 65,084
  • Somehow this is much more satisfying than any other proof. I am not quite sure why, but I think most mathematicians would agree that it is vastly superior to the hundreds of other proofs via chopping and rearranging. – Steven Gubkin Jul 13 '15 at 22:11
0

I would say Garfield's proof, which shows and two different ways to represent the area of the trapezoid. In that site it is the last one. Plus, the guy was one of our presidents at one point!