I say that the circle ABCD is to the circle EFGH as the square
on BD is to the square on FH.
For, if the square on BD is not to the square on FH as the circle
ABCD is to the circle EFGH, then as the square on BD is to the square
on FH, the circle ABCD is either to some less area than the circle
EFGH, or to a greater area.
First, let it be in that ratio to a less area S.
Inscribe the square EFGH in the circle EFGH. Then the inscribed
square is greater than the half of the circle EFGH, for if through
the points E, F, G, and H we draw tangents to the circle, then the
square EFGH is half the square circumscribed about the circle, and
the circle is less than the circumscribed square, hence the inscribed
square EFGH is greater than the half of the circle EFGH. IV.6
III.17
Bisect the circumferences EF, FG, GH, and HE at the points K, L,
M, and N. Join EK, KF, FL, LG, GM, MH, HN, and NE.
Therefore each of the triangles EKF, FLG, GMH, and HNE is also greater
than the half of the segment of the circle about it, for if through
the points K, L, M, and N we draw tangents to the circle and complete
the parallelograms on the straight lines EF, FG, GH, and HE, then
each of the triangles EKF, FLG, GMH, and HNE is half of the parallelogram
about it, while the segment about it is less than the parallelogram,
hence each of the triangles EKF, FLG, GMH, and HNE is greater than
the half of the segment of the circle about it. III.17
Thus, by bisecting the remaining circumferences and joining straight
lines, and by doing this repeatedly, we shall leave some segments
of the circle which will be less than the excess by which the circle
EFGH exceeds the area S.
For it was proved in the first theorem of the tenth book that if
two unequal magnitudes are set out, and if from the greater there
is subtracted a magnitude greater than the half, and from that which
is left a greater than the half, and if this is done repeatedly,
then there will be left some magnitude which is less than the lesser
magnitude set out. X.1
Let segments be left such as described, and let the segments of
the circle EFGH on EK, KF, FL, LG, GM, MH, HN, and NE be less than
the excess by which the circle EFGH exceeds the area S.
Therefore the remainder, the polygon EKFLGMHN, is greater than the
area S.
Now inscribe in the circle ABCD the polygon AOBPCQDR similar to
the polygon EKFLGMHN.
Therefore the square on BD is to the square on FH as the polygon
AOBPCQDR is to the polygon EKFLGMHN. XII.1
But the square on BD is to the square on FH as the circle ABCD to
the area S, therefore the circle ABCD is to the area S as the polygon
AOBPCQDR is to the polygon EKFLGMHN. Therefore, alternately the
circle ABCD is to the polygon inscribed in it as the area S is to
the polygon EKFLGMHN. V.11
V.16
But the circle ABCD is greater than the polygon inscribed in it,
therefore the area S is also greater than the polygon EKFLGMHN.
But it is also less, which is impossible.
Therefore the square on BD is to the square on FH not as the circle
ABCD is to any area less than the circle EFGH.
Similarly we can prove that the circle EFGH is to any area less
than the circle ABCD not as the square on FH is to the square on
BD.
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.
For, if possible, let it be in that ratio to a greater area S. Therefore,
inversely the square on FH is to the square on DB as the area S
is to the circle ABCD.
But the area S is to the circle ABCD as the circle EFGH is to some
area less than the circle ABCD, therefore the square on FH is to
the square on BD as the circle EFGH is to some area less than the
circle ABCD, which was proved impossible. Therefore the square on
BD is to the square on FH not as the circle ABCD to any area greater
than the circle EFGH. Lemma
V.11
And it was proved that neither is it in that ratio to any area less
than the circle EFGH, therefore the square on BD is to the square
on FH as the circle ABCD is to the circle EFGH.
Q.E.D.
LEMMA
I say that, the area S being greater than the circle EFGH the area
S is to the circle ABCD as the circle EFGH is to some area less
than the circle ABCD.
For let it be contrived that the area S is to the circle ABCD as
the circle EFGH to the area T.
I say that the area T is less than the circle ABCD.
Since the area S is to the circle ABCD as the circle EFGH is to
the area T, therefore, alternately the area S is to the circle EFGH
as the circle ABCD is to the area T. V.16
But the area S is greater than the circle EFGH, therefore the circle
ABCD is also greater than the area T.
Hence the area S is to the circle ABCD as the circle EFGH is to
some area less than the circle ABCD.
Therefore, circles are to one another as the squares on their diameters.
Q.E.D.
