## 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 $a$ $b$, 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 $c$ $a^2+b^2=c^c$.

(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.)