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BOOK XII.

LEMMA.

If from any given magnitude there be taken a magnitude greater than its half, and from the remainder a magnitude greater than its half, and so on, there shall at length remain a magnitude less than any other given magnitude less than the former.

Let AB and C be the two given magnitudes, of which AB is the greater: if from AB there be taken a magnitude greater than its half, and from the remainder a magnitude greater than its half, and so on, A there shall at length remain a magnitude less than C.

K

F

H

с

D

BCE

Take of C such a multiple DE as shall be greater than AB; and let DE be divided into parts, DF, FG, GE, each equal to C: From AB take BH, greater than its half, and from the remainder AH take HK greater than its half, and so on, till the number of the magnitudes BH, HK, KA is equal to the number of the magnitudes DF, FG, GE: Then, because DE is greater than AB, and that EG taken from DE is not greater than its half, but BH taken from AB is greater than its half, therefore the remainder DG is greater than the remainder AH: And, in like manner, DF is greater than AK: but DF is equal to C; therefore also C is greater than AK, that is, AK is less than C.

Wherefore, If from any given magnitude &c. Q. E.D. N.B. If only the halves be taken away, the same thing may in the same way be demonstrated.

PROP. I. THEOR.

Similar polygons inscribed in circles are to one another as the squares of their diameters.

Let the similar polygons ABCDE, FGHKL, be inscribed in circles, whose diameters are BM, GN: the polygon ABCDE shall be to the polygon FGHKL as the square of BM to the square of GN.

Join AM, BE, FN, GL: Then, because the polygons are similar, therefore the

angle BAE is equal to

the angle GFL, and BA B

is to AE as GF to FL: Therefore the two trian

gles BAE, GFL, having

A

[blocks in formation]

one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, are equiangular, and therefore the angle AEB is equal to the angle FLG: But the angle AEB is equal to the angle AMB, because they stand upon the same circumference, and, for the same reason, the angle FLG is equal to the angle FNG; therefore the angle AMB is equal to the angle FNG: And the right angle BAM is equal to the right angle GFN; therefore the triangles ABM, FGN are equiangular, and therefore BA is to GF as BM is to GN: Therefore also (5. Def. 10, and 5. 22) the duplicate ratio of BA to GF is the same with that of BM to GN: But (6. 20) the polygon ABCDE is to the polygon FGHKL in the duplicate ratio of BA to GF, and the square of BM is to the square of GN in the duplicate ratio of BM to GN; therefore the polygon ABCDE is to the polygon FGHKL as the square of BM to the square of GN.

Wherefore, Similar polygons &c.

Q. E. D.

PROP. II. THEOR.

Circles are to one another as the squares of their diameters.

Let ABCD, EFGH, be two circles, BD, FH, their diameters: the circle ABCD shall be to the circle EFGH as the square of BD to the square of FH.

For, if not, the square of BD must be to the square of FH as the circle ABCD is to some space either less than the circle EFGH, or greater than it: First, if possible, let it be to a space S, which is less than the circle EFGH; and in this circle inscribe the square

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EFGH: Then this square will be greater than half of the circle EFGH: (For, if through the points E, F, G, H, there be drawn lines touching the circle, it will be manifest, by joining EG, that the square EFGH is half of the square described about the circle, and therefore is greater than half of the circle :) Bisect the circumferences EF, FG, GH, HE in the points K, L, M, N, and join EK, KF, FL, LG, GM, MH, HN, NE: Then each of the triangles EKF, FLG, GMH, HNE, is greater than half of the segment upon the same base : (For, if through the points K, L, M, N, there be drawn lines touching the circle, and the parallelograms upon the lines EF, FG, GH, HE, be completed, each of the triangles will be half of the parallelogram upon the same base, and therefore greater than half of the segment upon the same base :) Therefore the sum of these triangles will be together greater than half the

sum of the segments, that is, than half the remainder of the circle, when the square EFGH is taken from it: Let these triangles also be taken away; then, by bisecting the circumferences EK, KF, &c., and forming triangles, as before, and taking them away, and so on continually, there will at length remain segments of the circle, which (Lemma) shall be together less than any given magnitude less than the circle EFGH, and therefore shall be together less than the excess of the circle EFGH above the space S: Let the segments EK, KF, &c. be those which remain, and are together less than the excess of the circle EFGH above the space S; then the rest of the circle, that is, the polygon EKFLGMHN, is greater than the space S.

Now, if in the circle ABCD be described a polygon AXBOCPDR similar to the polygon EKFLGMHN, then (x11. 1) the first polygon is to the second as the square of BD to the square of FH, that is, (Hyp.) as the circle ABCD is to the space S: But the first polygon AXBOCPDR is manifestly less than the circle ABCD, in which it is inscribed; therefore the second polygon, EKFLGMHN, is also less than the space S: But it has been proved to be greater than it—which is impossible: Therefore the square of BD is not to the square of FH as the circle ABCD is to any space less than the circle EFGH; and so also, in like manner, the square of FH is not to the square of BD as the circle EFGH to any space less than the circle ABCD.

Next, if possible, let the square of BD be to the square of FH as the circle ABCD to a space T, greater than the circle EFGH: Then, inversely, the square of FH is to the square of BD as the space T to the circle ABCD: But as the space T is to the circle ABCD so is the circle EFGH to some space, which must be less than

the circle ABCD, (since (Hyp.) the space T is greater than the circle EFGH;) therefore the square of FH is to the square of BD as the circle EFGH is to some space less than the circle ABCD—which has been just shewn to be impossible: Therefore the square of BD is not to the square of FH as the circle ABCD to any space greater than the circle EFGH, nor, as has been shewn before, to any space less than the circle EFGH; therefore the space of BD must be to the square of FH as the circle ABCD to the circle EFGH.

Wherefore, Circles are to one another as the squares of their diameters. Q. E.D.

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