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N arch of a circle is greater than its chord; but it is less Book XII. than two straight lines drawn from the same point, to touch the circle in the extremities of the arch.

PROP. I. THEOR *.

F from the greater of two unequal magnitudes,

there be taken more than its half, and from the remainder more than its half; and so on : There shall at length remain a magnitude less than the least of the proposed magnitudes. Let AB and C be two unequal magnitudes, of which AB is

If from AB there be taken more than its half, and from the remainder more than its half, and so on; there shall at length remain a 'magnitude less than C.

For may be multiplied so, as at length to become greater than AB; let it be so multiplied, and let DE its multiple be

greater * This is Prop. I. Book X. of Euclid.

the greater.

Book XII. greater than AB, and let DE be divided into 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, until there be as many di-

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D vifions in AB as there are in DE: and let the divisions in AB be AK, KH, HB; and the divisions in DE be DF, FG, GE: 'and be. K cause 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

Н. half; therefore the remainder GD is

greater than the remainder HA: Again, because GD

G is greater than HA, and that GF is not greater than the half of GD, but HK is greater than the half of HA; therefore the remainder FD is greater than the remainder AK : and FD B C is equal to C; therefore C is greater than AK; that is, AK is less than C. Q. E. D.

And if only the halves be taken away, the same thing may in the same way be demonstrated.

PROP. II. THEOR.

Е.

See N. VERY circle is equal to the rectangle contained

by its radius, and a straight line equal to the half of its circumference.

Let ABCD be a circle, of which the diameter is AC, and the

centre E; and in AC produced, let EF be taken equal to the a 11. 1. half of the circumference, and draw the diameter BD at right

angles to AC, and complete the rectangle DF; the rectangle DF
is equal to the circle ABCD.
If not, it is either greater or less than the circle.

First, let DF be greater than the circle; then some rectangle less than DF is equal to the circle ; let this be the rectangle DH: and divide

BE into such equal parts EK, KL, LM, MB, that each of them 1. 12. may be less than the fourth part of HFb; and draw" KN, LO,

9. 6. MP parallel to AC, meeting the circumference in N, O, P, and c 31. 1. complete the rectangle KČ, EN, NL, KO, OM, LP, PB and à 36. 1. because KE is equal to KL, the rectangle EN is equal d 10 NL:

For the fame reason, KO is equal to OM, and LP to PB; there fore the rectangles NL, OM, PB are equal to the rectangles EN, KO, LP in the circle; and the rectangle KC is therefore the ex. cess of the rectangles about the circle above those within it: and if the same construction be made in the other three sectors, it

may

b

may be proved in the same way, that all the circumscribed rect- Book XII.
angles exceed all the inscribed ones by four times the rectangle
KC: but because FH is greater than four times EK, the rect-
angle HG is greater than four times KC; and HD being equal to
the circle, is greater than the inscribed rectangles ; therefore DF
is greater than the inscribed rectangles, together with four times
KC; that is, greater than the circumscribed rectangles. Draw e 17. 3.
PY touching the circle ; and join PE; then the triangles EPY,
EBY together are equal to

f41. 1. half the rectangle contained

Z
by EB and BY, YP toge-
ther: But BY, YP are great-
er than the arch BP; there-

g Ax. 12.
fore the figure EBYP, and
much more EBXP, is greater

E
than the rectangle contained
by EB and the half of the
arch BP: In the same mana
ner, it may be proved, that
the quadrilateral EOVP is
greater than the rectangle
contained by EB and the half
of the arch PO, and so on :
Therefore all the circumscri.
bed rectangles together are
greater than the rectangle
contained by the radius and
the half of the circumfe.
rence ; that is, greater than
the rectangle DF: and it has
been proved, that they are B M L K

E

D
also less; which is impossible:
Therefore the rectangle DF
is not greater than the circle. Y
Neither is it lefs : For, if

W
poffible, let the circle be e XP

T qual to some rectangle DZ

VO

R

SN greater than DF: Then, if

С the same construction be made, it may be proved, as before, that the excess of the circumscri. bed rectangles above the inscribed ones is less than GZ: and the circumscribed rectangles are greater than the circle ; that is, than DZ; therefore the inscribed rectangles are greater than DF: Join BP; and the triangle EBP is less than the rectangle f 41. I. contained by EB and the half of BP; and BP is less & than the g Ax. 12. arch BP; and the triangle EMP is less than EBP; therefore

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Book XII. much more is the triangle EMP less than the rectangle contained

by EB and the half of the arch BP. In the same manner, it may be proved, that the quadrilateral EPWO is less than the rectangle contained by EP and the half of the arch PO; and so on: Therefore all the inscribed rectangles are less than the rectangle contained by the radius and the half of the circumfe. rence ; that is, less than DF: and they were proved to be greater than it ; which is impoflible ; therefore the rectangle DF is not less than the circle: and it was before proved, not to be greater than it; therefore DF is equal to the circle. Q. E. D.

Cor. 1. Hence it is manifest, that if a space be less than a circle, rectangles can be inscribed in the circle, which together are greater than that space : and that if a space be

a fpace be greater than a circle, rectangles can be described about the circle, which are together less than that space.

Cor. 2. And consequently, if a figure be greater than any series of rectangles in a circle, but less than any series of rect. angles about it; that figure is equal to the circle.

PROP. III. THEOR *.

CFR

IRCLES are to one another as the squares of their

diameters. Let ABCD, EFGH be two circles, and BD, FH their diame. ters: as the square of BD to the square of FH, so is the circle ABCD to the circle EFGH.

Let the square of BD be to the square of FH, as the circle

ABCD to some space Z; this space is equal to the circle EFGH. a 11. 1. Draw a the diameters AC, EG at right angles to BD, FH; and 6 9.6. divide AK into any number of equal parts in M, N, O, and

EL into as many in P, Q, R ; and complete the rectangles KS,
MT, NV; LW, PX, QY in the circles ; and let the same con-

struction be made in the rest of the sectors: and because AO, € 15. 5. ER are the same parts of OC, RG; AO is to ER, as COC to

RG; therefore the similar rectilineal figures upon them are prod 22. 6. portionals d: and the squares of A0, ER are similar figures on e 1. Def.6. them; and the rectangles AO, OC and ER, RG are similar

figures on OC, RG; therefore the square of AO is to the square of ER, as the rectangle AO, OC to the rectangle ER,

RG: and AO is equal to ON, and ER to RQ; and the rectf 35. 3. angle AO, OC is equalf to the square of OV, and the rectangle ER, RG to the square of RY; therefore the square of ON is to

that

* This is Prop. II. Book XII, of Euclid.

that of RQ, as the square of OV to that of RY; therefore NO Book XII. is to RO as OV to RY; and, by alternation, NO is to OV, as QR to RY; therefore the rectangle NV is similare to QY; & 16.5. therefore NV is to YQ, as the square of NO to the square of RQ; that is, as the square of BD to the square of FH °, be- C 15. 5. caufe ON, RQ are the same parts of BD, FH. In the same manner,

it

may be proved, that the other rectangles are to one another as the square of BD to that of FH: Wherefore all the

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2. 12.

rectangles in the circle ABCD are to all the rectangles in the circle EFGH as the square of BD to the square of FH: But h 12. 5. as the square of BD to the square of FH, so is the circle ABCD to the space Z; therefore, as the circle ABCD is to the space Z, so * are the rectangles in the circle ABCD to the rectangles in k 11. s. the circle EFGH: and the circle ABCD is greater than the rectangles in it; therefore the space Z is greater than the rectangles q 14. 5. in the circle EFGH. In the same manner, it may be proved, that the space Z is less than any series of rectangles about the circle EFGH: Therefore the space Z is equal to the circle 1 2. Cor. EFGH: But the square of BD is to the square of FH, as the circle ABCD to the space Z; therefore the square of BD is to that of FH, as the circle ABCD to the circle EFGH. Q. E. D.

Cor. Hence, the circumferences of circles are to one another as their diameters.

For a circle is equal to the rectangle contained by the radius m 2. 12. and half the circumference; therefore the rectangle contained by BK and half the circumference ABCD is to the rectangle contained by FL and half the circumference EFGH, as o the circle

n 7. 5. ABCD to the circle EFGH; that is, as the square of BD to o 3. 12. that of FH, or as the square of BK to that of FL; and, alternately 5, as the rectangle contained by BK and half the circumference ABCD is to the square of BK, so is the rectangle contained by FL and half the circumference EFGH to the square of FL: and parallelograms of the same altitude are as their bases p ; P 1. 6. therefore the half of the circumference ABCD is to BK, as half the circumference EFGH to FL ; and, alternately 8, as half the

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