And the angle BAM is equal to the angle GFN, for each of them is a right angle. [III. 31.
Therefore the remaining angles in the triangles AMB, FNG are equal, and the triangles are equiangular to one another;
therefore BA is to BM as GF is, to GN, [VI. 4.
and, alternately, BA is to GF as BM is to GN, [V. 16.
therefore the duplicate ratio of BA to GF is the same as the duplicate ratio of BM to GN. [V. Definition 10, V. 22.
But the polygon ABCDE is to the polygon FGHKL in the duplicate ratio of BA to GF; [VI. 20.
and the square on BM is to the square on GN in the duplicate ratio of BM to GN; [VI. 20.
therefore the polygon ABCDE is to the polygon FGHKL as the square on BM is to the square on GN [V. 11.
Wherefore, similar polygons &c. q.e.d.
PROPOSITION 2. THEOREM.
Circles are to one another as the squares on their diameters.
Let ABCD, EFGH be two circles, and BD, FH their diameters: the circle ABCD shall be to the circle EFGH as the square on BD is to the square on FH.
For, if not, the square on BD must be to the square on FH as the circle ABCD is to some space either less than the circle EFGH, or greater than it.
First, if possible, let it be as the circle ABCD is to a space S less than the circle EFGH.
