If this process be continued, and the triangles be supposed to be taken away, there will at length remain segments of circles which are together less than the excess of the circle EFGH above the space S, by the preceding Lemma.
Let then the segments EK, KF, FL, LG, GM, MH, HN, NE be those which remain, and which are together less than the excess of the circle above S;
therefore the rest of the circle, namely the polygon EKFLGMHN, is greater than the space S.
In the circle ABCD describe the polygon AXBOCPDR similar to the polygon EKFLGMHN.
Then the polygon AXBOCPDR is to the polygon EKFLGMHN as the square on BD is to the square on FH, [XII. 1.
that is, as the circle ABCD is to the space S. [Hyp., V. 11.
But the polvgon AXBOCPDR is less than the circle ABCD in which it is inscribed,
therefore the polygon EKFLGMHN is less than the space S; [V. 14.
but it is also greater, as has been shewn;
which is impossible.
Therefore the square on BD is not to the square on FH as the circle ABCD is to any space less than the circle EFGH
In the same way it may be shewn that the square on FH is not to the square on BD as the circle EFGH is to any space less than the circle ABCD.
Nor is the square on BD to the square on FH as the circle ABCD is to any space greater than the circle EFGH.
For, if possible, let it be as the circle ABCD is to a space T greater than the circle EFGH.
Then, inversely, the square on FH is to the square on BD as the space T is to the circle ABCD.
But as the space T is to the circle ABCD so is the circle EFGH to some space, which must be less than the circle
