We start with a right-angled triangle with sides labeled $a,b$ and $c\text{.}$ Then we construct a square of side-length $a+b$ and draw inside it 4 copies of the triangle arranged as shown in the centre of the above figure. The area in white is then $a^2+b^2\text{.}$ Now move the triangles around to create the arrangement shown on the right of the above figure. The area in white is bounded by a square of side-length $c$ and so its area is $c^2\text{.}$ The area of the outer square didn't change when the triangles were moved, nor did the area of the triangles, so the white area cannot have changed either. This proves $a^2+b^2=c^2\text{.}$