PROPOSITION 2 BOOK XII

Proposition 2. Circles are to one another as the squares on their diameters. Let ABCD and EFGH be circles, and let BD and FH be their diameters.

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.

 Copyright Applet © 1996/1997 (Juny, 1997) D.E.Joyce Clark University © Drets d´ús cedits 2002/2003 JDL The thirteen books of Euclid's Elements translated from the text of Heiberg with introduction and commentary. Copyright © Thomas Little Heath, 1908