Construct a line from B perpendicular to AC and extend it to the bottom of the square ACGF.

Begin with a right triangle ABC.

The idea of the proof is to "slide" triangles about the figure, always parallel to their bases. In that way the areas do not change.

 

Construct the square upon the hypotenuse AC.

Then construct the square upon the leg AB

 

We will show that the square ABDE constructed upon the leg AB has the same area as the rectangle ALMF.

This is the first half the proof.

The second half of the proof will be to show that the area of the square upon the leg BC is the same as the area of the rectangle LCGM.

Half the area under consideration is the triangle ADB, whose area we will show is half the rectangle ALMF.

This proof is essentially identical to the proof originally in Euclid's The Elements.

It requires less "geometrical" machinery than any of the other proofs, many of which rely on similarity.

Slide the vertex B of triangle ABD parallel to its base AD to the point C.

These triangles have the same area.

Rotate the triangle ADC (90 degrees) to become the triangle ABF.

Again the area is preserved.

Now slide the vertex B parallel to the base AF to the point M. The area remains invariant. Thus, half the original area of ADEB is half the area of the rectangle ALFM.

The second half of the proof is to repeat the process showing that the square upon the leg BC has the same area at the rectangle LCMG.