It is thought that the great Pythagoras and his students derived the first proof that in a right triangle, the sum of the squares of the perpendicular sides is equal to the square of its longest side (hypotenuse). Through the years numerous derivation had come up to prove this theorem.

Below are the list of some of the proofs of the Pythagorean Theorem.

**Proof #1**

Start with the given triangle in this manner,

drop a line from the point of intersection of** a** and **b **to line **c,**

giving us two smaller right triangles with the hypotenuse equal to **a** and **b** which are similar to the bigger triangle. Thus giving us,

**x = b ² / c
**

**(c – x) = a² / c
**

**h = ab / c**

The area of the big right triangle can be determined by adding the area of the two right triangle,

**(xh)/2 + h*(c – x)/2 = ab/2**

substituting **x, c – x** and **h**, we will have,

**ab³ + a³b = abc²**

**a² + b² = c²**

**Proof #2
**

We start with 4 copies of our given triangle and arrange them in this manner:

One right triangle has an area of **1/2ab.**

Now let’s put them altogether to form a square with side equal to **c**. An inner square can also be formed with a side equal to **a – b.**

The area of the bigger square is **c²** and is equal to the sum of the areas of the four right triangles and the inner square which will give us,

**c² = 4(1/2ab) + (a – b)²**

**c² = 2ab + (a ² – 2ab + b²)**

**c² = a ² + b²**

**Proof #3**

This proof was first published by James Garfield, the 20th U.S. President.

Construct two same right triangles arranged in this manner:

Construct a figure using these two triangles and draw a line from one vertex of the first triangle to the other vertex of the second triangle. This will give us a trapezoid with two parallel bases one equal to **a** and the other equal to **b**. One side is vertical (the trapezoid’s altitude) **a+ b** and the other side is equal to **c.**

Solving for the area of the trapezoid, we will have **(a+b)/2 * (a+b). **Looking at the trapezoid in another way, we can solve it by adding the three right triangles, **(ab)/2 + (ab)/2 + (c²)/2**. Equating these two areas thus gives us,

**(a+b)/2 * (a+b)** = **(ab)/2 + (ab)/2 + (c²)/2**

**a² + b² + 2ab = 2ab + c²**

**c² = a² + b²
**

For more fun derivations of the Pythagorean Theorem visit:

http://www.cut-the-knot.org/pythagoras/index.shtml

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