Easy foorp

That proof can also be done sort of backwards and is arguably easier that way. Instead of using three triangles of known dimensions and proving they can be assembled into a rectangle, start with the rectangle and find the dimensions of the triangles.

Take a right triangle with sides ab, and c. Draw perpendiculars to the hypotenuse at each end and a parallel line through the third vertex to make a rectangle BCDE.

Angles CDA and EAD are congruent (alternate interior angles). So are ADE and ACD since both are complementary with ADC. So triangles ADE and ACD are similar; the hypotenuses are in the ratio b/c, so the vertical leg DE = a(b/c) and the horizontal leg AE = b(b/c). Likewise we find BC = b(a/c) (as it should be) and AB = a(a/c). But then BE = CD implies a(a/c)+b(b/c) = c or, multiplying though by ca^2+b^2=c^c.

(Right away you know this proof fails in non Euclidean geometry, because in the first steps it relies on there being one and only one line parallel to CD through A.)


Easy proof

Sheesh, how have I lived this long and not seen this proof of the Pythagorean Theorem?

Here’s a right triangle, with legs a and b and hypotenuse c.Here’s the same triangle scaled up by a factor of a, so the legs are a^2 and ab and the hypotenuse is ac.And here it is twice more, scaled up by factors of b… and c.Of course these all are similar triangles.

Now, notice the hypotenuse of the first scaled-up triangle is the same length as one leg of the third, so we can draw them juxtaposed like this:Capital letters label the vertices.

Since the triangles are similar, angles ADC and ACB are congruent, and since they’re right triangles, angles ADC and ACD add to a right angle; therefore angle BCD is a right angle.

Similarly the hypotenuse of the second scaled triangle is congruent to the other leg of the third, so they can be juxtaposed as well:and as before, angle CDE is a right angle. Also, angles DAE and BAC sum to a right angle so angle BAE is two right angles, i.e.B, A, and E are colinear. Therefore BCDE is a rectangle, and BE = CD. But BE = a^2+b^2 and CD = c^2. So a^2+b^2=c^2. QED.

What I like about this proof is how easy it is to grasp visually. Most proofs of the Pythagorean Theorem deal with areas, specifically persuading you the areas of two squares sum to the same as the area of a third square, but summing areas isn’t easy to visualize. Here there are no areas, just lengths, and we all can see immediately that a^2+b^2 is just BE.

(I saw this proof on, um, some blog recently. And failed to make note of exactly where. Sorry.)