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Topic: [HM] "President Garfield's Proof"
Replies: 20   Last Post: Feb 29, 2000 8:46 PM

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Clark Kimberling

Posts: 20
Registered: 12/3/04
Re: [HM] "President Garfield's Proof"
Posted: Jan 15, 2000 10:03 AM
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An HM reader has asked which proof of the Pythagorean theorem is the
one attributed to Garfield. The answer may be of general interest.

"Garfield's proof" goes with a figure consisting of three right
triangles. Let ABC be the given one, with hypotenuse BA and shortest
side BC. Draw right triangle ABD with length(BD)=length(BA) and D on
the side of AB that doesn't contain C. Then draw right triangle BDE
with length(DE)=length(BC) and E on the side of BD that doesn't
contain A. Triangle BDE is congruent to triangle ABC. Let a,b,c
denote the common sidelengths. To finish the proof, I'll quote
Malcolm Graham's article, "President Garfield and the Pythagorean
Theorem," in The Mathematics Teacher, Dec. 1976 (in a series for the
American Bicentennial called "Events in the History of American

"In figure 1, we see a trapezoid with bases a and b and height (a+b).
The trapezoid is the union of three right triangles. Hence, the area
of the trapezoid is equal to the sum of the areas of the three

(a+b)(a+b)/2 = ab/2 + ab/2 + (c^2)/2
(a+b)(a+b) = ab ab + c^2
a^2 + 2ab + b^2 = 2ab + c^2
a^2 + b^2 = c^2 ."

In Elisha Scott Loomis's collection, The Pythagorean Proposition
(National Council of Teachers of Mathematics, 1968) this proof is on
page 231 and is also number 231 in the collection (a fixed-point
theorem caught in action!)

Eric Weisstein's remarkable CRC Concise Encyclopedia of Mathematics,
1999, on page 1465, includes "A novel proof ... discovered by James
Garfield." I mention this to indicate that the "Garfield proof" is
gaining in popularity. It has also been reproduced in other recent

Ah - even better: the same author's Mathworld ("The Web's Most
Extensive Math Resource") offers the same coverage. Mathworld is at . From there you can reach the "Garfield
proof" by typing Garfield into the search box. Or, you can go
directly to and
scroll down aways.

Clark Kimberling

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