A proof of the Pythagorean Theorem

A 3-4-5 triangle

The Pythagorean Theorem states that the area of the square on the hypotenuse of a right-angled triangle equals the sum of the areas of the squares on the other two sides. Algebraically (where c is the length of the hypotenuse and a and b are the lengths of the other two sides):

    c² = a² + b²

I thought it would be fun to attempt to prove the theorem for myself. This post describes my solution.

Figure 1: Draw a right-angled triangle with squares

1. Draw a right-angled triangle with squares

• Draw a right-angled triangle
• Draw the squares extending from each side
• Label the sides a, b and c.

Figure 2: Flip the square on the hypotenuse

2. Flip the square on the hypotenuse

• Flip the square on the hypotenuse over
• Mark dotted lines to indicate the larger square containing the flipped square [1]

Figure 3: Label the segments

3. Label the segments

• Mark the segments on each side of the dotted square and label as follows:
• Label the top and right segments a and b as guided by the red and green squares
• Mark the top segment on the left side b, since it is the third side of a right-angled triangle with a and hypotenuse c as the other two sides
• Mark the bottom segment on the left side a, since it is symmetrical to the right side
• Mark the right segment on the bottom side a, since it is the third side of a right-angled triangle with b and hypotenuse c as the other two sides
• Mark the left segment on the bottom side b, since it is symmetrical to the top side

4. Do some algebra

• The area of the square on the hypotenuse equals the area of the large square minus the areas of the triangles in the four corners:

    c² = (a + b)(a + b) - 4(a * b / 2)
       = a² + b² + 2ab - 2ab
       = a² + b²


Which is the Pythagorean Theorem!

It turns out that there are over a hundred possible proofs. The above proof is similar to proof #4 which is credited to the 12th century Hindu mathematician Bhaskara II.

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Figure 4: Side and corner collinearity

[1] Note that the outer sides of the red and green squares are on the dotted lines. That each outer side is collinear with its adjacent yellow square corner is shown by rotating and shifting the original blue triangle to construct the sides of the flipped yellow square.

The square in the center has a side length of (b - a). As an alternative proof, the area of the square on the hypotenuse equals the area of the center square plus the area of the four triangles:

    c² = (b - a)(b - a) + 4(a * b / 2)
       = b² + a² - 2ab + 2ab
       = a² + b²