Book 12
2
Let segments be left such as described, and let the segments of the circle EFGH on EK, KF, FL, LG, GM, MH, HN, NE be less than the excess by which the circle EFGH exceeds the area S.
2
Therefore the remainder, the polygon EKFLGMHN, is greater than the area S.
2
Let there be inscribed, also, in the circle ABCD the polygon AOBPCQDR similar to the polygon EKFLGMHN; therefore, as the square on BD is to the square on FH, so is the polygon AOBPCQDR to the polygon EKFLGMHN. [XII. 1]
2
But, as the square on BD is to the square on FH, so also is the circle ABCD to the area S; therefore also, as the circle ABCD is to the area S, so is the polygon AOBPCQDR to the polygon EKFLGMHN; [V. 11] therefore, alternately, as the circle ABCD is to the polygon inscribed in it, so is the area S to the polygon EKFLGMHN. [V. 16]
2
But the circle ABCD is greater than the polygon inscribed in it; therefore the area S is also greater than the polygon EKFLGMHN.
2
But it is also less: which is impossible.
2
Therefore, as the square on BD is to the square on FH, so is not the circle ABCD to any area less than the circle EFGH.
2
Similarly we can prove that neither is the circle EFGH to any area less than the circle ABCD as the square on FH is to the square on BD.
2
I say next that neither is the circle ABCD to any area greater than the circle EFGH as the square on BD is to the square on FH.
2
For, if possible, let it be in that ratio to a greater area S.
2
Therefore, inversely, as the square on FH is to the square on DB, so is the area S to the circle ABCD.
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.
