James Garfield Pythagorean Theorem (Illustration & Proof)

Mr. James Abram Garfield proposed this proof in 1876. The surprising part is that Mr. Garfield was a politician! And not an ordinary one, he became the 20th president of the US. Wow!

Mr. James Abram Garfield, US President who proved the Pythagoras Theorem

He proved the Pythagoras theorem with the help of geometrical construction and formulas for the area of triangle and trapezium. So you should already know the following formulas to understand the proof:

Area of a Triangle = $\frac{1}{2} \times base \times altitude$
Area of a Trapezium = $\frac{1}{2} \times (Sum of parallel sides) \times (distance between them)$

Any grade 8 student should be able to understand this proof. Let us see how he did it!

Statement of Pythagoras Theorem

Pythagoras theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.

AB² + AC² = BC²

As AB = a, AC = b and BC = c, we have to prove:

a² + b² = c²

Proof

Step 1 – A Simple Construction

We create a copy of the right triangle and rotate it by 90 degrees in the clockwise direction.

Construction step one. The triangle is rotated clockwise by 90 degrees

Next, we move up the triangle to align line segments AB and BD. So we form a straight line segment AD.

Construction step two. The triangle is shifted up so that AB and BD are aligned

That’s it!

Step 2 – Find angle ∠5

Find the angle between two triangles

In the triangle ABC:

∠1 + ∠2 + ∠3 = 180

∠1 + 90 + ∠3 = 180

∠1 + ∠3 = 180 – 90 = 90

As AD is straight line segment, we can say:

∠1 + ∠5 + ∠3 = 180

∠5 + ∠1 + ∠3 = 180

We can replace ∠1 + ∠3 by 90

∠5 + 90 = 180

∠5 = 90

We will use this later in our proof. Next we join E and C to form trapezium ADEC ( We know that DE is parallel to AC. How? )

Step 3 – Area of Trapezium ADEC

Area of Trapezium ADEC calculated using the formula

The formula for the area of trapezium is:
$\frac{1}{2} \times (Sum of parallel sides) \times (distance between them)$

Here parallel sides are DE and AC. So sum of parallel sides = a + b.

Distance between the parallel sides is AD = AB + BD = a + b. Using these values in the formula we get:

Area $= \frac{1}{2} \times (a + b) \times (a + b)$

$= \frac{1}{2} \times (a^{2} + 2 ab + b^{2})$

Step 4 – Area of Trapezium ADEC from Triangles

Area of Trapezium ADEC calculated as the sum of the area of triangles

Three right triangles – ABC, BDE, and BCE form the trapezium. So its area is the sum of areas of these triangles.

Area of Trapezium = area of triangle ABC + area of triangle BCE + area of triangle BDE

$= \frac{1}{2} \times (a . b) + \frac{1}{2} \times (c . c) + \frac{1}{2} \times (a . b)$

$= \frac{1}{2} \times (c^{2} + 2 ab)$

Notice that we used ∠5 = 90 degrees in calculating the area of the triangle BCE.

Step 5 – Equating the Areas

In the end, we equate the areas obtained in the previous two steps.

$\frac{1}{2} \times (a^{2} + 2 ab + b^{2}) = \frac{1}{2} \times (c^{2} + 2 ab)$

Multiplying both sides by 2:

$2 \times \frac{1}{2} \times (a^{2} + 2 ab + b^{2}) = 2 \times \frac{1}{2} \times (c^{2} + 2 ab)$

a² + 2ab + b² = c² + 2ab

Subtracting 2ab from both sides, we get:

a² + b² = c²

We have proved the theorem!

There are many ways to prove the Pythagoras Theorem, however we chose one of the simplest methods.

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