Proof of Garfield's Pythagorean theorem

In addition to the "classical" proofs of the Pythagorean theorem , such as geometric proof by supplementation , shear proof or proof with similarities , a proof was developed by James A. Garfield around 1875 and submitted to the New England Journal of Education and even published. James A. Garfield became President of the United States of America in 1881 .

Proof from James A. Garfield

Evidence sketch

A right triangle is given (see graphic). ${\displaystyle \Delta ABC}$

By shifting along and rotating around with an angle of 90 ° you get a triangle . The two triangles are congruent : ${\displaystyle \Delta ABC}$ ${\displaystyle {\overline {BA}}}$ ${\displaystyle A}$ ${\displaystyle \Delta A\!\,''B\!\,''C\!\,''}$

${\displaystyle \Delta ABC\cong \Delta A\!\,''B\!\,''C\!\,''.}$

From the congruence theorems it follows:

${\displaystyle a=a^{\prime \prime },\;b=b^{\prime \prime },\;c=c^{\prime \prime }.}$

According to the inside angle sum theorem in the triangle:

${\displaystyle \alpha +\beta +\gamma =180^{\circ }}$.

It follows with : ${\displaystyle \gamma =90^{\circ }}$

${\displaystyle \alpha +\beta =90^{\circ }.}$

Furthermore, since the angle is stretched (180 °) and is, it follows ${\displaystyle \angle CAC^{\prime \prime }}$ ${\displaystyle \alpha +\beta =90^{\circ }}$

${\displaystyle \angle A^{\prime \prime }B^{\prime \prime }B=90^{\circ }.}$

So all three triangles are right triangles. Their surface area is calculated from half the product of the leg lengths ( ). ${\displaystyle A_{{\text{rechtwinkliges }}\Delta }={\tfrac {1}{2}}\cdot a\cdot b}$

By drawing in the route you get a trapezoid as a geometric figure . Its area is calculated using the formula ${\displaystyle {\overline {BA^{\prime \prime }}}}$ ${\displaystyle A_{\text{Trapez}}={\tfrac {1}{2}}(a+b^{\prime \prime })\cdot h.}$

From the equality of area it follows that the area of ​​the trapezoid corresponds to the sum of the areas of the three triangles:

${\displaystyle {\begin{matrix}A_{\text{Trapez}}&=&A_{{\text{rotes }}\Delta }+A_{{\text{blaues }}\Delta }+A_{{\text{grünes}}\;\Delta }&{}&{}\\{\frac {1}{2}}(a+b)(a+b)&=&{\frac {1}{2}}ab+{\frac {1}{2}}ab+{\frac {1}{2}}cc&|&{\text{binomische Formel}}\\{\frac {1}{2}}(a^{2}+2ab+b^{2})&=&ab+{\frac {1}{2}}c^{2}&|&\cdot \,2\\a^{2}+2ab+b^{2}&=&2ab+c^{2}&|&-2ab\\a^{2}+b^{2}&=&c^{2}&{}&{}\end{matrix}}}$

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• JA Garfield, Pons Asinorum. New England J. Educ. 3, p. 161, 1876.
• Eric Weisstein: Pythagorean Theorem . In: MathWorld (English). Formulas 19-24.