PROPOSITION 2.
Circles are to one another as the squares on the diameters.
Let
ABCD,
EFGH be circles, and
BD,
FH their diameters; I say that, as the circle
ABCD is to the circle
EFGH, so is the square on
BD 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, so will the circle
ABCD be either to some less area than the circle
EFGH, or to a greater.
First, let it be in that ratio to a less area
S.
Let the square
EFGH be inscribed in the circle
EFGH; then the inscribed square is greater than the half of the circle
EFGH, inasmuch as, if through the points
E,
F,
G,
H we draw tangents to the circle, 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.
Let the circumferences
EF,
FG,
GH,
HE be bisected at the points
K,
L,
M,
N, and let
EK,
KF,
FL,
LG,
GM,
MH,
HN,
NE be joined; therefore each of the triangles
EKF,
FLG,
GMH,
HNE is also greater than the half of the segment of the circle about it, inasmuch as, if through the points
K,
L,
M,
N we draw tangents to the circle and complete the parallelograms on the straight lines
EF,
FG,
GH,
HE, each of the triangles
EKF,
FLG,
GMH,
HNE will be half of the parallelogram about it, while the segment about it is less than the parallelogram; hence each of the triangles
EKF,
FLG,
GMH,
HNE is greater than the half of the segment of the circle about it.
Thus, by bisecting the remaining circumferences and joining straight lines, and by doing this continually, 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 be set out, and if from the greater there be subtracted a magnitude greater than the half, and from that which is left a greater than the half, and if this be done continually, there will be left some magnitude which will be less than the lesser magnitude set out.
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.
Therefore the remainder, the polygon
EKFLGMHN, is greater than the area
S.
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]
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]
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, 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.
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.
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, as the square on
FH is to the square on
DB, so is the area
S to the circle
ABCD.
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.
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.
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.
Therefore etc. Q. E. D.
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.
For let it be contrived that, as the area
S is to the circle
ABCD, so is the circle
EFGH to the area
T.
I say that the area
T is less than the circle
ABCD.
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]
But the area
S is greater than the circle
EFGH; therefore the circle
ABCD is also greater than the area
T.
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.