New-England Journal of Education, 1876

The Congressman's Trapezoid

James A. Garfield · 1876 · AlgebraicDifficultyClassical

In a right triangle with legs $a$ , $b$ and hypotenuse $c$ , the identity $a^{2} + b^{2} = c^{2}$ falls out of computing one trapezoid's area by two different methods.

Place two copies of a right triangle with legs $a$ and $b$ so that one's leg $b$ and the other's leg $a$ lie along a single straight line — together making a segment of length $a + b$ .
Join their two free corners with a straight segment. That closes a trapezoid: parallel sides of length $a$ and $b$ , standing $a + b$ apart.
The two segments running from the meeting point out to the trapezoid's slanted corners are each a hypotenuse of length $c$ — and the angle between them is a right angle, since the triangle's two acute angles are complementary and sit side by side here.
So the trapezoid is really three triangles: the original triangle, a mirrored copy of it, and a right isosceles triangle with legs $c$ .
Write the trapezoid's area two ways — once by the trapezoid formula, once as the sum of the three triangles — and set them equal: $\frac{1}{2} (a + b)^{2} = \frac{1}{2} ab + \frac{1}{2} ab + \frac{1}{2} c^{2}$ .
Expand and cancel the matching $2 ab$ terms on each side, and what's left is $a^{2} + b^{2} = c^{2}$ .

Therefore c² = a² + b². ∎

James Garfield found this while serving in the House of Representatives, five years before his presidency; it ran in the New-England Journal of Education credited only to “General James A. Garfield, M. C.” It is one of the few well-known proofs that never draws a square.