Book 12
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.
PROPOSITION 3.
3
Any pyramid which has a triangular base is divided into two pyramids equal and similar to one another, similar to the whole and having triangular bases, and into two equal prisms; and the two prisms are greater than the half of the whole pyramid.
3
Let there be a pyramid of which the triangle ABC is the base and the point D the vertex; I say that the pyramid ABCD is divided into two pyramids equal to one another, having triangular bases and similar to the whole pyramid, and into two equal prisms; and the two prisms are greater than the half of the whole pyramid.
3
For let AB, BC, CA, AD, DB, DC be bisected at the points E, F, G, H, K, L, and let HE, EG, GH, HK, KL, LH, KF, FG be joined.
3
Since AE is equal to EB, and AH to DH, therefore EH is parallel to DB. [VI. 2]
3
For the same reason HK is also parallel to AB.
3
Therefore HEBK is a parallelogram; therefore HK is equal to EB. [I. 34]
3
But EB is equal to EA; therefore AE is also equal to HK.
