James Garfield's Pythagorean Proof

Today I learned James Garfield, who once worked as a lawyer, Civil War General, and served as the 20th President of the United States, was math savvy and published a novel Pythagorean theorem proof.^[1]

\[ \text{Area}_{\text{trapezoid } ACED} = \frac{1}{2} \cdot (AC + DE) \cdot CE = \frac{1}{2} \cdot (a + b) \cdot (a + b) = \frac{(a + b)^2}{2} \] \[ \begin{aligned} \text{Area}_{\text{trapezoid } ACED} &= \text{Area}_{\Delta ACB} + \text{Area}_{\Delta ABD} + \text{Area}_{\Delta BDE} \\ &= \frac{1}{2}(a \times b) + \frac{1}{2}(c \times c) + \frac{1}{2}(a \times b) \end{aligned} \] \[ (a + b) \times \frac{1}{2}(a + b) = \frac{1}{2}(a \times b) + \frac{1}{2}(c \times c) + \frac{1}{2}(a \times b) \] \[ a^{2} + b^{2} = c^{2} \]

Small Pieces

We can take this in smaller pieces. First, we can find the area of the right-angled trapezoid with the following equation:

\[ \text{Area}_{\text{trapezoid}} = \frac{1}{2} \cdot (a + b) \cdot (a + b) = \frac{(a + b)^2}{2} \]

We can find the area of each of the two outer triangles with the following:

\[ \text{Area}_{\text{triangle}} = \frac{ab}{2} \]

And the area of the inner triangle with:

\[ \text{Area}_{\text{inner triangle}} = \frac{c^2}{2} \]

Proof

Reducing, we can go to the end, beginning with our substituted and now simplified area equation demonstrated above:

\[ \frac{(a + b)^2}{2} = 2 \cdot \frac{ab}{2} + \frac{c^2}{2} \]

Then we expand \( (a + b)^2 \) on the left hand side. And our equation on the right can also be simplified since we're both multiplying and dividing \( ab \) by 2:

\[ \frac{a^2 + 2ab + b^2}{2} = ab + \frac{c^2}{2} \]

Multiply both sides by 2 to eliminate denominators:

\[ a^2 + 2ab + b^2 = 2ab + c^2 \]

Lastly, subtract \( 2ab \) from both sides:

\[ a^2 + b^2 = c^2 \]

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Footnotes

1. Mathematical Treasure: James A. Garfield’s Proof of the Pythagorean Theorem ↩︎