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The Pythagorean Theorem states that the square of the length of the hypotenuse of any right triangle equals the sum of the squares of the lengths of the legs.

The action figure below gives a wonderful illustration of a geometric proof of the Pythagorean Theorem.

Experiment with the action figure below by sliding the small blue squares in the left hand figure along the axes.

Below is the algebraic explanation of how this figure shows that u2 + v2 = w2, where u and v label the legs of the triangle on the left, and w is the hypotenuse.

In order to talk about the different parts of this figure, let's give some labels to the right-hand square of the action figure above. These labels are shown in Figure 1.	FIGURE 1
Consider this square, ABCD, with side lengths u + v. Locate P on side AB so that |AP| = v and |PB| = u. In the same way, locate Q, R and S on the sides BC, CD and DA respectively.
We see that <APB, <BQC, <CRD and <DSA are straight angles. That is, they measure 180o. We want to show that <SPQ, <PQR, <QRS and <RSP are each right angles in order to show that the interiour quadrilateral, PQRS, is a square. (By construction, we already know all four sides of this quadrilateral have the same length, w.)
Notice that SAP and PBQ each have one side of length u, one side of length v and that the enclosed angle in each case is 90o. Thus, by the Side-Angle-Side theorem, these two triangles are congruent, that is, identical except for location.
Thus <APS + <QPB = 90o. This implies that <SPQ = 180o - (<APS + <QPB) = 180o - 90o = 90o.
In the same way we can show that QCR, RDS, SAP and PBQ are all congruent. It then follows that each of the angles in the quadrilateral PQRS is 90o, thus PQRS is a square.
Now we compute the area of ABCD in two ways to finally show that u2 + v2 = w2. This will prove the Pythagorean Theorem.
Method 1: Area(ABCD) = (length of side)2 = (u + v)2 = u2 + 2uv + v2 = u2 + v2 + 4(uv/2)	Method 2: Area(ABCD) = 4(area of green triangle) + (area of interior square) = 4(uv/2) + w2
Because the area of ABCD computed by either method must be the same, we can equate these two expressions and arrive at : u2 + v2 + 4(uv/2) = 4(uv/2) + w2 ==> u2 + v2 = w2.