Pythagorean Theorem proofs
The Pythagorean Theorem states that, in a right triangle, the square of a (a²) plus the square of b (b²) is equal to the square of c (c²):


a² + b² = c² 
Many different proofs exist for this most fundamental of all geometric theorems.
Algebraic proof. Based on four equal right triangles.
We start with a right triangle (in gold, in the diagram) with sides a, b, and c.
We then build a big square, out of four copies of our triangle. We end up with a square, in the middle, with sides c, as shown.
It is a big square, with each side having a length of a+b, so the total area is (a+b)^{2}
Now let's add up the areas of all the smaller pieces:
 First, the smaller (tilted) square has an area of c^{2}
 There are four triangles, each one has an area of A = ab, so all four of them combined is 4()ab=2ab
 So, adding up the tilted square and the 4 triangles gives A = c^{2}+2ab
The area of the large square is equal to the area of the tilted square and the 4 triangles. This can be written as:
(a+b)(a+b) = c²+2ab
NOW, let us rearrange this to see if we can get the pythagoras theorem:
Start with: 

(a+b)(a+b) 
= 
c² + 2ab 





Expand (a+b)(a+b): 

a² + 2ab + b² 
= 
c² + 2ab 





Subtract "2ab" from both sides: 

a² + b² 
= 
c² 
concluding Pythagoras' proof.
Proof by subtraction:
In this proof, the square on the hypotenuse plus four copies of the triangle can be assembled into the same shape as the squares on the other two sides plus four copies of the triangle.
The area of the first square is given by (a+b)^{2} or 4(ab)+ a^{2} + b^{2}.
The area of the second square is given by (a+b)^{2} or 4(ab) + c^{2}.
Since the squares have equal areas we can set them equal to another and subtract equals. The case (a+b)^{2}=(a+b)^{2} is not interesting. Let's do the other case.
4(ab) + a^{2} + b^{2} = 4(ab)+ c^{2}
Subtracting equals from both sides we have a² + b² = c², concluding Pythagoras' proof.