Showing that the two rhomboids are equal to the two rectangles.
Demonstration B: Based on Equivalence. In this second demonstration the relative equivalence of the rhomboid, the rectangles, and the squares is shown outside the figure by means of the parallel indentures which are on both sides of the frame. These indentures, when the pieces are placed in them, show that the pieces have the same altitude.
This is the manner of procedure: Starting again with the original position, take out the two rectangles and place them in the parallel indentures to the left, the larger in the wider indenture and the smaller in the narrower indenture. The different figures in the same indenture have the same altitude; therefore the pieces need only to be placed together at the base to prove that they are equal—hence the figures are equal in pairs: the smaller rectangle equals the smaller rhomboid and the larger rectangle equals the larger rhomboid.
Starting again from the original position you proceed analogously with the squares. In the parallel indentures to the right the large square may be placed in the same indenture with the large rhomboid, which, however, must be turned in the opposite direction (in the direction of its greatest length); and the smaller square and the smaller rhomboid fit into the narrower indenture. They have the same altitude; and that the bases are equal is easily verified by putting them together; therefore here is proof that the squares and the rhomboids are respectively equivalent.
Rectangles and squares which are equivalent to the same rhomboids are equivalent to each other. Hence the theorem is proved.
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Showing that the two rhomboids are equal to the two squares.