Garfield's Proof of the Pythagorean Theorem: the Congressman's Trapezoid

A proof from Congress

On 1 April 1876, the New-England Journal of Education, a teachers' weekly published in Boston, printed a short proof of the Pythagorean theorem. Its author was not a mathematician. He was a member of the United States House of Representatives — and five years later he became President.

His proof is still taught, because it needs almost nothing: three right triangles and the area of a trapezoid.

Who was Garfield?

James Abram Garfield was born in Ohio in 1831. Before politics he was a teacher, and later the head of the Western Reserve Eclectic Institute in Hiram, Ohio. He served as a general in the Civil War, and from 1863 he sat in the House of Representatives for seventeen years.

He was elected the 20th President in 1880. In July 1881, only four months into his term, he was shot; he died in September. His proof dates from 1876, when he was still a congressman.

What the journal printed

The note carried the title Pons Asinorum and was signed with the initials J. A. G. The editor introduced it like this:

Pons asinorum — "the bridge of asses" — is traditionally the name of a different result, Euclid's Proposition I.5 about the base angles of an isosceles triangle. Here the name was used for the most famous theorem of all.

Building the trapezoid

1. Take a right triangle with legs a and b and hypotenuse c. Lay it on its leg b, so that its leg a stands upright.
2. Next to it, lay a second copy on its leg a, so that its leg b stands upright. The two bottom legs now form one straight line of length a + b.
3. Join the two upper corners. The outline is a trapezoid: two parallel, upright sides a and b, at a distance a + b from each other.

The middle triangle has two sides of length c — the hypotenuses of the two copies. Why is the angle between them a right angle? Along the straight bottom line, the angles at the meeting point add up to 180°. The two outer ones are the two acute angles of the right triangle, and those add up to 90°. That leaves exactly 90° in the middle.

Counting the area twice

First, as a trapezoid. Area = half the sum of the parallel sides × their distance:

½ · (a + b) · (a + b)

Second, as three triangles. The two blue triangles each have area ab ⁄ 2; the red one has area c² ⁄ 2:

ab⁄2 + ab⁄2 + c²⁄2

Both describe the same shape, so they are equal. Double both sides and multiply out:

a² + 2ab + b² = 2ab + c²

The 2ab cancels on both sides, and what is left is the theorem:

a² + b² = c²

Half of an older picture

Arrange four copies of the right triangle inside a square of side a + b, leaving a smaller tilted square of side c in the middle. Arrangements like this belong to a long tradition of picture proofs of the theorem.

Cut the big square along a diagonal of the tilted square, and each half is exactly Garfield's trapezoid. Geometrically, his proof is one half of this larger picture.

Hundreds of proofs

Garfield's proof is one of hundreds. In 1927, the Ohio teacher Elisha Scott Loomis published a collection of proofs of the theorem; an expanded edition appeared in 1940 and includes Garfield's. Some are by famous mathematicians, some by students — and one by a future President.

For the classical proof with squares on all three sides, see Euclid's proof (Elements I.47).

Try it yourself

Take the triangle with legs 5 and 12 and hypotenuse 13.

• The trapezoid has parallel sides 5 and 12 and width 17: area ½ · 17 · 17 = 144.5.
• The two outer triangles: 30 each. The middle triangle: ½ · 13 · 13 = 84.5.
• Check: 30 + 30 + 84.5 = 144.5.