1.) Start with a right triangle whose sides are a and b and with hypotenuse c ==>


2.) Make a large square consisting of 4 of the right triangles with an inscribed square whose sides are c in length:


3.) Now, the sides of the larger square are (a + b) in length. The area of this big square is:

Big square area = (a + b)^{2}

4.) The area of this large square is also the sum of the areas of the 4 right triangles (which is ab/2) plus the area of the inner square, which is c^{2}:

Big square area = 4 (ab/2) + c^{2}

5.) Since these 2 areas are the same we can set them equal to each other:

4 (ab/2) + c^{2} = (a + b)^{2}

6.) By the distributive property: (a + b)^{2} = (a + b) (a + b) = a (a + b) + b ( a + b) = a^{2} + ab + ab + b^{2} = a^{2} +2ab + b^{2}
 Putting this back in gives us ===>


4 (ab/2) + c^{2} = a^{2} + 2 ab + b^{2}

7.) Simplifying the 4(ab/2) term, we have:

2ab + c^{2} = a^{2} + 2ab + b^{2}

8.) Finally, subtracting 2ab from both sides of the equation gives us the theorem of Pythagoras: 
2ab + c^{2} = a^{2} + 2ab + b^{2}
2ab 2ab
