Pythagoras' Theorem

Now a parallel is drawn to the hypotenuse passing by the center of the square of the longest side of the triangle. The intersection of this line with the opposing sides of that square determines the shaded parallelogram in the figure.

Through the same square's center another straight line, perpendicular to the longest side of the parallelogram, is drawn. With that, the longest side square's area is divided into four equal rectangle trapezoids, therefore with the same area each.

Now if the trapezoids are laid out within the area of the square the side of which is the triangle's hypotenuse, a small square remains in the center. The side of this small square is exactly equal to the difference between sides "a" and "c" of each trapezoid, which in turn is exactly equal to the triangle's smallest side, marked "b" in the figure (in green).

Result: the sum of the squares (by their areas) of the sides of the rectangle triangle is equal to the square of the triangle's hypotenuse, quod erat demonstrandum!