Garfield’s Proof of The Pythagorean
Garfield’s Proof
James A. Garfield was the twentieth president of the USA. He studied at The Williams College and later taught in the Western Reserve Electric Institute after his graduation. Other ventures which he was involved in include, practicing law, serving as the president of the Western Reserve, and election to the United States Congress. Garfield was also privileged to come up with an original proof of the Pythagorean Theorem. He developed this proof in 1876 while serving in the Congress (Sid, 2016). This makes in the New-England journal of education dated April 1, 1876. The editor of the journal deliberately called the theorem Pons Asinorum or “Bridge Asses,” which was a nickname given to the isosceles triangle theorem
Garfield’s proof consists of a trapezoid with bases “a” and “b” and height “a + b”. He viewed the area of the trapezoid in two different ways first as a trapezoid and secondly as three right angle triangles of which two of them are congruent. This theorem gained more popularity because the previous theorems were difficult to understand which was bridged by this theorem by Garfield.
In this proof, you start with a right triangle of sides “a, b”, and “c”, where “c” is the hypotenuse. Then you are required to extend length a by b and form an identical triangle along the extension. Now we have two duplicate triangles which are joined at their apex as shown in the figure above. From the triangles, it is true to say that the lower leg b is parallel to the upper leg a. joining point X to Y complete the trapezoid construction which consists of three triangles of which two of them are replicas. The other triangle XYZ is an isosceles triangle because of the two lengths c.
Another task which we need to know is the size of the angle XZY where the two side c’s come across. Following the laws of a triangle which states that the sum of the angle within a triangle is equal to 180 degrees, we can deduce that the angle XZY=90 degrees.
This finding confirms that this trapezoid is formed by three right triangles.
The next step is finding the area of the trapezoid from individual triangles, then later find the area using the trapezoidal formula. The area of the two identical triangles if formulated as 2*1/2(a*b) = a*b (after multiplying 2 by 1/2). We also represent the area of the third triangle as ½*c*c. the sum of the two findings makes the area of the trapezoid which is equal to a*b + 1/2c^2. From the formula for a trapezoid this are can be represented as (area = height * (base1 + base2)/2) where the height is equal to a + b and (base1 + base2)/2 is equal to ½(a + b) therefore the final area = ½ (a +b) ^2.
The two areas arrived at above are equal {[1/2 (a + b) ^2] = [(a*b) +1/2c^2]}. Next we multiply both sides by 2 two to eliminate the 1/2s from the equation, this simplifies this equation further to. (a + b)^2 = 2ab + c^2. Expanding the first part of the equation gives (a +b) ^2 = a^2 + b^2 + 2ab. Replacing this in the initial equation produces, a^2 + b^2 +2ab = 2ab + c^2. Subtracting 2ab from both sides leads us to a conclusion that a^2 +b^2 = c^2 thus proofing the Pythagorean Theorem
The algebraic relationships are summarized in the table bellow
| Area of the trapezoid | = | 2(area of the scalene right triangles | + | Area of the isosceles triangle |
| ½ ( a + b)(a + b) | = | 2(1/2ab) | + | ½ c^2 |
| (a + b) (a +b) | = | 2ab | + | c^2 |
| a^2 + 2ab + b^2 | = | 2ab | + | c^2 |
| a^2 + b^2 | = | c^2 |
References
Sid, J. K. (2016). Mathematical Treasure: James A. Garfield’s Proof of the Pythagorean. Mathematical Association of America, 10.
