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.
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
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
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.