to P (15. 5.), because MN and NP are equimultiples of M and P and for the same reason, the ratio of NP to PQ is the same with that of N to Q; therefore the ratio of MN to PQ is compounded of the ratios of M to P, and of N to Q. Now, the ratio of M to P is the same with that of the side AB to the side DE (by Hyp.); and the ratio of N to Q the same with that of the side BC to the side EF. Therefore, the ratio of MN to PQ is compounded of the ratios of AB to DE, and of BC to EF. And the ratio of the parallelogram AC to the parallelogram DF is compounded of the same ratios (23. 6.); therefore, the parallelogram AC is to the parallelogram DF as MN, the product of the numbers M and N, to PQ, the product of the numbers P and Q. Therefore &c. Q. E. D.

Cor. 1. Hence, if GH be to KL as the number M to the number N; the square described on GH

will be to the square described on G

KL as MM,the square of the num

ber M to NN, the square of the number N.

COR. 2. If A, B, C, D, &c. are any lines, and m, n, r, s, &c. numbers proportional to them; viz. A: B::m:n: A: C :: m : r, A ; Dms, &c.; and if the rectangle contained by any two of the lines be equal to the square of a third line, the product of the numbers proportional to the first two, will be equal to the square of the number proportional to the third; that is, if A.C-Ba,m Xr=nXn, pr=n?:

For by this Prop. A.C: B2 :: mXr : n2 ; but A.C=B3, therefore m X r=n. Nearly in the same way it may be demonstrated, that whatever is the relation between the rectangles contained by these lines, there is the same between the products of the numbers proportional to them.

So also conversely if m and r be numbers proportional to the lines A and C; if also A C—B2, and if a number n be found such, that n2 mr, then A: B:: m; n. For let A B mq, then since, m, 4, are proportional to A, B, and C, and A.C=B2; therefore, as has just been proved, qa=mxr; but n2=qxr, by hypothesis, therefore ng", and n=q; wherefore A: B::m: n.

SCHOLIUM.

In order to have numbers proportional to any set of magnitudes of the same kind, suppose one of them to be divided into any number, m of equal parts, and let H be one of those parts. Let H be found »

times in the magnitude B, r times in C, s times in D, &c., then it is evident that the numbers m, n, r, s are proportional to the magnitudes A, B, C and D. When therefore it is said in any of the following propositions, that a line as A = a number m, it is understood that A=m XH, or that A is equal to the given magnitude H multiplied by m; and the same is understood of the other magnitudes, B, C, D, and their proportional numbers, H being the common measure of all the magnitudes. This common measure is omitted for the sake of brevity in the arithmetical expression; but is always implied, when a line, or other geometrical magnitude, is said to be equal to à number. Also, when there are fractions in the number to which the magnitude is called equal, it is meant that the common measure H is farther subdivided into such parts as the numerical fraction indicates. Thus, if A=360.375, it is meant that there is a certain magnitude H, 375 such that A=360×H+- XH, or that A is equal to 360 times H, 1000

together with 375 of the thousandth parts of H. true in all other cases, where numbers are used to tions of geometrical magnitudes,

And the same is express the rela

The perpendicular drawn from the centre of a circle on the chord of any arch is a mean proportional between half the radius and the line made up of the radius and the perpendicular drawn from the centre on the chord of double that arch: And the chord of the arch is a mean proportional between the diameter and a line which is the difference between the radius and the foresaid perpendicular from the centre.

Let ADB be a circle, of which the centre is C; DBE any arch, and DB the half of it; let the chords DE, DB be drawn ; as also CF and CG at right angles to DE and DB; if CF be produced it will meet the circumference in B: let it meet it again in A, and let AC be bisected in H; CG is a mean proportional between AH and AF; and

BD a mean proportional between AB and BF, the excess of the radius above CF.

Join AD; and because ADB is a right angle, being an angle in a simicircle; and because CGB is also a right angle, the triangles ABD, CBG are equiangular, and, AB: AD :: BC: CG (4. 6.), or alternately, AB BC: AD CG; and therefore, because AB is double of BC, AD is double of CG, and the square of AD therefore equal to four times the square of CG.

But, because ADB is a right angled triangle, and DF a perpendicular on AB, AD is a mean proportional between AB and AF (8. 6.), and AD2=AB.AF (17. 6.), or since AB is =4AH, AD2=4ÀH.AF. Therefore also, because 4CG2=AD2, 4CG24AH.AF, and CG2= AH.AF; wherefore CG is a mean proportional between AH and AF, that is, between half the radius and the line made up of the radius, and the perpendicular on the chord of twice the arch BD.

Again, it is evident, that BD is a mean proportional between AB and BF (8. 6.), that is, between the diameter and the excess of the radius above the perpendicular, on the chord of twice the arch DB. Therefore, &c. Q. E. D.

PROP. IX. THEOR.*

The circumference of a circle exceeds three times the diameter, by a line less than ten of the parts, of which the diameter contains seventy but greater than ten of the parts whereof the diameter contains seventy one. Let ABD be a circle, of which the centre is C, and the diameter AB; the circumference is greater than three times AB, by a line

10 1

10

less than or of AC, but greater than of AC.

70' 7'

71

In the circle ABD apply the straight line BD equal to the radius

*In this proposition, the character + placed after a number, signifies that something is to be added to it; and the character-, on the other hand, signifies that something is to be taken away from it.

:

BC Draw DF perpendicular to BC, and let it meet the circumfer ence again in E; draw also CG perpendicular to BD produce BC to A, bisect AC in H, and join CD.

It is evident, that the arches BD, BE are each of them one-sixth of the circumference (Cor. 15. 4), and that therefore the arch DBE is one third of the circumference. Wherefore, the line (8. 1. Sup) CG is a mean proportional between AH, half the radius, and the line AF. Now because the sides BD, DC, of the triangle BDC are equal, the angles DCF, DBF are also equal; and the angles DFC, DFB being equal, and the side DF common to the triangles DBF, DCF, the base BF is equal to the base CF, and BC is bisected in F.

Therefore, if AC or BC=1000, AH=500, CF=500, AF=1500, and CG being a mean proportional between AH and AF, CG2=(17. 6.) AH.AF=500 X 1500 750000; wherefore CG=866.0254+, because (866.0254) is less than 750000. Hence also,AC+CG=1866. 0254+.

Now, as CG is the perpendicular drawn from the centre C, on the chord of one-sixth of the circumference, if P = the perpendicular from C on the chord of one-twelfth of the circumference, P will be a mean proportional between AH (8. 1. Sup.) and AC+CG, and P2 = AH (AC+CG)= 500 X (1866.0254+)=933012.7+. Therefore, P=965.9258+, because (965.9258) is less than 933012.7. Hence also, AC+P=1965,9258+.

Again, if the perpendicular drawn from C on the chord of one twenty-fourth of the circumference, Q will be a mean proportional between AH and AC+P, and Q=AH (AC+P)=500(1965.9258+) =982962.9+; and therefore Q-991.4449+, because (991.4449) is less than 982962.9. Therefore also AC+Q=1991.4449+.

In like manner, if S be the perpendicular from C on the chord of one forty-eighth of the circumference, S2 AH (AC+Q) = 500 (1991.4449+)=995722.45+; and S 997.8589+, because (997 8589) is less than 995722.45. Hence also, AC+S=1997.8589+. Lastly, if T be the perpendicular from C on the chord of one ninety-sixth of the circumference, T2, AH (AC+S)=500 (1997.8589 +) 998929.45+, and T 999.46458+. Thus T, the perpendicular on the chord of one ninety-sixth of the circumference, is greater than 999.46458 of those parts of which the radius contains 1000.

=

But by the last proposition, the chord of one ninety-sixth part of the circumference is a mean proportional between the diameter and the excess of the radius above S, the perpendicular from the centre on the chord of one forty-eighth of the circumference. Therefore, the square of the chord of one ninety-sixth of the circumference AB (AC-S) 2000 × (2,1411—,) = 4282.2—; and therefore the chord itself 65.4386-, because (65.4386)2 is greater than 4282.2. Now, the chord of one ninety-sixth of the circumference, or the side of an equilateral polygon of ninety-six sides inscribed in the circle,

=

being 65.4386, the perimeter of that polygon will be (65.4386 -)96

=6282.1056-.

Let the perimeter of the circumscribed polygon of the same num ber of sides, be M, then (2. Cor. 2. 1. Sup.) T: AC :: 6282.1056— . M, that is, (since T 999.46458+, as already shewn),

999.46458+: 1000 :: 6282.1056-: M; if then N be such, that 999.46458: 1000 :: 6282.1056-: N; ex æquo perturb. 999.46458 +: 999.46458 :: N: M; and, since the first is greater than the second, the third is greater than the fourth, or N is greater than M.

Now, if a fourth proportional be found to 999.46458, 1000 and 6282.1056 viz. 6285.461-, then,

because, 909.46458 1000 :: 6282.1056: 6285.461—,

and as before, 999.46458: 1000 :: 6282.1056—: N;

therefore, 6282.1056: 6282.1056-:: 6285.461-N, and as the first of these proportionals is greater than the second, the third, viz. 6285/

461 is greater than N, the fourth. But N was proved to be greater than M; much more, therefore, is 6285.461 greater than M, the perimeter of a polygon of ninety-six sides circumscribed about the circle ; that is, the perimeter of that polygon is less than 6285.461; now, the circumference of the circle is less than the perimeter of the polygon; much more, therefore, is it less than 6285.461; wherefore the circumference of a circle is less than 6285.461 of those parts of which the radius contains 1000. The circumference, therefore, has to the diameter a less ratio (8. 5.) than 6285.461 has to 2000, or than 3142.7305 has to 1000 but the ratio of 22 to 7 is greater than the ratio of 3142.7305 to 1000, therefore the circumference has a less ratio to the diameter than 22 has to 7, or the circumference is less than 22 of the parts of which the diameter contains 7.