Garfield's proof of the Pythagorean theorem is an original proof of the Pythagorean theorem discovered by James A. Garfield, the 20th president of the United States. The proof appeared in print in the New-England Journal of Education (Vol. 3, No. 14, April 1, 1876). At the time of the publication of the proof, Garfield was a congressman from Ohio. He assumed the office of President in 1881, and served in that position until his death later that year after being shot in an assassination. Garfield is thus far the only President of the United States to have contributed anything original to mathematics. The proof is nontrivial and, according to the historian of mathematics William Dunham, "Garfield's is really a very clever proof." The proof appears as the 231st proof in The Pythagorean Proposition, a compendium of 370 different proofs of the Pythagorean theorem.
The proof
In the figure, A B C {\displaystyle ABC} is a right-angled triangle with right angle at C {\displaystyle C} . The side-lengths of the triangle are a , b , c {\displaystyle a,b,c} . Pythagorean theorem asserts that c 2 = a 2 + b 2 {\displaystyle c^{2}=a^{2}+b^{2}} . To prove the theorem, Garfield drew a line through B {\displaystyle B} perpendicular to A B {\displaystyle AB} and on this line chose a point D {\displaystyle D} such that B D = B A {\displaystyle BD=BA} . Then, from D {\displaystyle D} he dropped a perpendicular D E {\displaystyle DE} upon the extended line C B {\displaystyle CB} . From the figure, one can easily see that the triangles A B C {\displaystyle ABC} and B D E {\displaystyle BDE} are congruent. Since A C {\displaystyle AC} and D E {\displaystyle DE} are both perpendicular to C E {\displaystyle CE} , they are parallel and so the quadrilateral A C E D {\displaystyle ACED} is a trapezoid. The theorem is proved by computing the area of this trapezoid in two different ways.
area of trapezoid A C E D = height × average of parallel sides = C E × 1 2 ( A C + D E ) = ( a + b ) × 1 2 ( a + b ) . {\displaystyle {\begin{aligned}{\text{area of trapezoid }}ACED&={\text{height}}\times {\text{average of parallel sides}}\\&=CE\times {\tfrac {1}{2}}(AC+DE)=(a+b)\times {\tfrac {1}{2}}(a+b).\end{aligned}}}
area of trapezoid A C E D = area of △ A C B + area of △ A B D + area of △ B D E = 1 2 ( a × b ) + 1 2 ( c × c ) + 1 2 ( a × b ) . {\displaystyle {\begin{aligned}{\text{area of trapezoid }}ACED&={\text{area of }}\triangle ACB+{\text{area of }}\triangle ABD+{\text{area of }}\triangle BDE\\&={\tfrac {1}{2}}(a\times b)+{\tfrac {1}{2}}(c\times c)+{\tfrac {1}{2}}(a\times b).\end{aligned}}}
From these one gets
( a + b ) × 1 2 ( a + b ) = 1 2 ( a × b ) + 1 2 ( c × c ) + 1 2 ( a × b ) {\displaystyle (a+b)\times {\tfrac {1}{2}}(a+b)={\tfrac {1}{2}}(a\times b)+{\tfrac {1}{2}}(c\times c)+{\tfrac {1}{2}}(a\times b)}
which on simplification yields
a 2 + b 2 = c 2 . {\displaystyle a^{2}+b^{2}=c^{2}.}
Relationship to other proofs
Garfield's proof is a variant of one of the algebraic proofs (pictured at right), but using only half of the diagram. The pictured version observes that the area of the outer square equals the area of the inner square plus four congruent triangles, which is to say
( a + b ) × ( a + b ) = ( c × c ) + 4 × 1 2 ( a × b ) {\displaystyle (a+b)\times (a+b)=(c\times c)+4\times {\tfrac {1}{2}}(a\times b)}
and simplifies the same way.
References