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Kitap 12

2 But, as the area S is to the circle ABCD, so is the circle EFGH to some area less than the circle ABCD; therefore also, as the square on FH is to the square on BD, so is the circle EFGH to some area less than the circle ABCD: [V. 11] which was proved impossible.

2 Therefore, as the square on BD is to the square on FH, so is not the circle ABCD to any area greater than the circle EFGH.

2 And it was proved that neither is it in that ratio to any area less than the circle EFGH; therefore, as the square on BD is to the square on FH, so is the circle ABCD to the circle EFGH.

2 Therefore etc. Q. E. D.

2 Lemma. I say that, the area S being greater than the circle EFGH, as the area S is to the circle ABCD, so is the circle EFGH to some area less than the circle ABCD.

2 For let it be contrived that, as the area S is to the circle ABCD, so is the circle EFGH to the area T.

2 I say that the area T is less than the circle ABCD.

2 For since, as the area S is to the circle ABCD, so is the circle EFGH to the area T, therefore, alternately, as the area S is to the circle EFGH, so is the circle ABCD to the area T. [V. 16]

2 But the area S is greater than the circle EFGH; therefore the circle ABCD is also greater than the area T.

2 Hence, as the area S is to the circle ABCD, so is the circle EFGH to some area less than the circle ABCD. Q. E. D.