From Euclid's Elements,
Proposition xii.2 1
(Source: The Thirteen Books of the Elements, trans. with intro.
and commentary by Sir Thomas L. Heath, 2nd ed., Dover, 1956, vol. 3, pp.
371--373)
Circles are to one another as
the squares on the diameters. 2
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. 3
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. 4
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 half
of the circle EFGH, inasmuch as, if through the points E,
F,
G,
H
we draw tangents to the circle, the square 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. 5
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. 6
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. 7
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 exceeds the area S. 8
Therefore the remainder, the polygon
EKFLGMHN,
is greater than the area S. 9
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].10
But, as the square on BD
is to the square on FH, so also is the circle ABCD to the
area S 11;
therefore also, as the circle ABCD is to the area
S, so is
the polygon AOBPCQDR to the polygon
EKFLGMHN [v.11]12;
therefore, alternately, as the circle
ABCD is to the the polygon
AOBPCQDR
inscribed in it, so is the area S to the polygon
EKFLGMHN
[v.16]. 13
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.14
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. 15
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. 16
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. 17
But, as the area S is to
the circle ABCD, so is the circle EFGH to some area less
than the circle ABCD 18;
therefore also, as the square on
FH is to the square on DB,
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.
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