Garfield’s proof of the Pythagorean theorem

Mathematical proof by 1876 US President

Garfield’s proof of the Pythagorean theorem The 20th President of the United States, James A. There is an original proof of the Pythagorean theorem, discovered by Garfield (November 19, 1831 – September 19, 1881). proof of this appeared in print New-England Journal of Education (Vol. 3, No. 14, April 1, 1876).^[1][2] At the time of the publication of the evidence, Garfield was a congressman from Ohio. He assumed the office of President on March 4, 1881, and served in that position until his death on September 19, 1881, having succumbed to gunshot wounds in an assassination in July.^[3] Garfield is the only President of the United States to have made any original contribution to mathematics. This proof is non-trivial and, according to the historian of mathematics William Dunham, “Garfield actually has a very clever proof.”^[4] The proof appears as the 231st proof in pythagorean propositionA collection of 370 different proofs of the Pythagorean theorem.^[5]

In the picture, ${displaystyle abc}$ is a right angled triangle whose angles are at right angles ${displaystyle c}$The lengths of the sides of the triangle are ${displaystyle a,b,c}$The Pythagorean theorem emphasizes that ${displaystyle c^{2}=a^{2}+b^{2}}$,

Garfield drew a line to prove the theorem ${displaystyle b}$ perpendicular to ${displaystyle AB}$ and chose a point on this line ${displaystyle D}$ such that ${displaystyle BD=BA}$Again ${displaystyle D}$ he dropped a vert ${displaystyle DE}$ on extended line ${displaystyle cb}$From the figure, one can easily see that the triangle ${displaystyle abc}$ And ${displaystyle bde}$ are congruent. since ${displaystyle ac}$ And ${displaystyle DE}$ both are perpendicular ${displaystyle ce}$They are parallel and hence quadrilateral ${displaystyle ACED}$ is a trapezoid. The theorem is proved by calculating the area of ​​this trapezium in two different ways.

${displaystyle {begin{align} text{area of ​​trapezoid}} ACED&={text{height}} times {text{average of parallel sides}}$,

${displaystyle {start {align} text {area of ​​trapezoid }}ACED&={text{area }}\Delta ACB+{\text{area }}\Delta ABD+{\text{area }}\Delta BDE\&={\tfrac {1}{2}}(a\times b)+{\tfrac {1}{2}}(c\times c)+{\tfrac {1}$

get from these

${displaystyle (a+b)\times {tfrac {1}{2}}(a+b)={\tfrac {1}{2}}(a\times b)+{\tfrac {1}{2}}(c\times c)+{\tfrac {1}{2}}(a\times b)}$

which is obtained on simplification

${displaystyle a^{2}+b^{2}=c^{2}}$