Four Proofs of the Pythagorean Theorem

To give the Pythagorean Theorem in geometric terms we start with a right triangle abc. From the sides of this triangle we will draw squares from each of the sides, using the lengths of the sides as the length of a side for each respective square. The Pythagorean Theorem then claims that the sum of the areas of the two small squares equals (the area of) the large one.

In algebraic terms, a2+b2=c2 where c is the hypotenuse while a and b are the sides of the triangle.

The Pythagorean Theorem is probably the most well-known mathematical proofs, which many people remember well after their last course in mathematics.

 

 

 

 

 

Proof #1

For this proof we start with two squares with sides a and b, respectively, placed side by side. The total area of the two squares is a2+b2.
After removing the segment common to the two squares we draw a line from the upper left-hand corner of square a to a point on the base. The length of the base of this new triangle is set to be equal to b. From the lower end of this new hypotenuse we will draw a line to the upper right-hand corner of what used to be square b. Since the sum of the lenghts of the two original squares was a+b and we had drawn the first triangle to have a base of length b, the longer side of the second triangle is obviously of length a.

From geometry we know that corresponding sides of congruent triangles are equal, thus we know that the length of each hypotenuse is the same and we will call this length c.

As a last step, we rotate the triangles 90o, each around its top vertex. The right one is rotated clockwise whereas the left triangle is rotated counterclockwise. The resulting shape is a square with the side c and area c2. Since this new area equals the area of the two squares a and b we started with, this proves that a2+b2=c2.

 

 

 

 

 

 

Proof #2

This proof was discovered by President James A. Garfield in 1876. As opposed to the previous proof, we draw no squares, but simply use algebra to solve for a2+b2=c2. The key is to use the formula for the area of a trapezoid - half sum of the bases times the altitude = (a+b)/2 * (a+b). We will put this formula for the area of the trapezoid on the left-hand side of the equation. We will then put the sum of the area of the three triangles on the right-hand side of the equation. After simplification we will then get back to a2+b2=c2.

 

 

 

 

 

 

 

 

 

 

 

 

Proof #3

We start with the original triangle, now denoted ABC, and need only one additional construct - the altitude AD.

We will then draw each of the resulting three triangles.




 

 

 

 

The triangles ABC, BDA and ADC are similar which leads to two ratios:


and



By cross-multiplying our ratios, we get
and

 

By adding the two ratios we get


Since the sides of our original triangle were AB, AC, and BC, this proves that the sum of the squares of the sides of a right triangle equal the square of the hypotenuse.



 

 

 

 

Proof #4

For this proof we start with right triangle abc. After that we copy the triangle three times and rotate each of those triangles by one quarter turn. Each has area ab/2. Let's put them together without additional rotations so that they form a square with side c.

The square has a square hole with the side (a-b). We set the area of the square equal to the area of the four triangles less the are of the square hole and then simplify to get to a2+b2=c2 .