For, by the proposition, A.C: B2 : : mxr: n2. But A.C Book I. =B2, therefore mrn. 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. Conversely, if m and r be numbers proportional to the lines A and C, and if A.C-B2, and if a number n be found such that n2=mr, then A: B:: m : n. Let A: B:: m : q, then, since m, q, r are proportional to A, B, C, and A.CB2; therefore, as has just been proved, q=mxr. But n2=mxr, by hypothesis, therefore n2=q, 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 the magnitudes to be divided into any number, m, of equal parts, and let H be one of those parts. Let H be found n 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, D. Therefore when it is said in any of the following propositions that a line as Aa number m, it is understood that A=mxH, 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 a number. Also, when there are fractions in the number to which the magnitude is said to be 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, such that A-360 H+37 H, or that A is equal to 360 times H, together with 375 of the thousandth parts of H. And the same is true in all other cases, where numbers are used to express the relations of geometrical magnitudes. Suppl. a 31. 3. b 4. 6. c 16. 5. d C. 5. e 2.Cor.8.2. f 8. 6. g 17. 6. PROP. X. THEOR. 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; let DBE be any arch, and DB the half of it; let the chords DE, DB be drawn; let CF and CG be drawn at right angles to DE and DB; let CF be produced both ways Join AD. Because ADB is a right an glea, and because CGB is also a right angle, the triangles ABD, CBG are equiangular; therefore AB: AD :: BC: CG; therefore AB: BC:: AD: CG. But AB is double of BC, therefore AD is double of CGd, therefore AD-4CG2. Because ADB is a right angled triangle, and DF a perpendicular on AB, AD is a mean proportional between AB and AFf, therefore AD-AB.AFs, or, since AB=4AH, AD2= 4AH.AF But 4CG2=AD2, therefore 4CG2-4AH.AF, or CG-AH.AF; wherefore CG is a mean proportional between AH and AF8, 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, BD is a mean proportional between AB and BFf, that is, between the diameter and the excess of the radius above the perpendicular on the chord of twice the arch DB. Therefore, the perpendicular, &c. Q. E. D. Book I. PROP. XI. 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 less than 48, or, of AB, but greater than In the circle ABD apply the straight line BD equal to the radius BC; draw DF perpendicular to BC, and produce it to meet the circumference in E; draw CG perpendicular to BD; bisect AC in H, and join CD. It is evident that each of the arches BD, BE is one sixth of the circum A of AB. * In this proposition the character + placed after a number signifies that something is to be added to it, and the character to be taken from it. signifies that something is Suppl. ference, therefore the arch DBE is one-third of the cir cumference. Wherefore the line CG is a mean proportional b Cor. 15.4. between AH, half the radius, and the line AF. Because the c10.1. Sup. sides BD, DC, of the triangle BDC, are equal, the angles DCF, DBF are also equal. In the triangles DBF, DCF the angles DFC, DFB are equal, and the angles DCF, DBF are equal, and the side DF is common to both; therefore the base BF is equal to the base CFd, therefore BC is bisected in F. d 26. 1. Now if AC-1000, then AH-500, CF-500, AF=1500, therefore CG2- AH.AF=500X1500-750000; wherefore CG=866.0254+, because (866.0254) is less than 750000. Hence AC+CG=1866.0254+. Because 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 proportionale between AH 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 AC+P=1965.9258+. If Q 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+; therefore Q=991.4449+, because (991.4449) is less than 982962.9. Hence AC+Q= 1991.4449+. 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+; therefore S=997.8589+,because (997.8589)2 is less than 995722.45. Hence AC+S= 1997.8589-+. Lastly, if T be the perpendicular from C on the chord of one-ninety-sixth of the circumference, TAH(AC+S)= 500(1997.8589+)=998929.45+; therefore 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. 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 part of the circumference; therefore the square of the chord of oneninety-sixth of the circumference AB (AC-S) = 2000× (2.1411-)=4282.2-; therefore the chord-65.4386-, be Book 1. cause (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—. f 23. 5. Let the perimeter of the circumscribed polygon of the same number of sides be M, then T: AC:: 6282.1056— M, e 2. Cor. or, (T=999.46458+), 999.46458+: 1000:: 6282.1056: M. 3. 1. Sup. Let N be such that 999.46458 : 1000:: 6282.1056—: N; ex æquo perturbate, 999.46458+: 999.46458 :: N: M; alternately, 999.46458+: N:: 999.46458 : M; therefore, since the first is greater than the third, the second is greater than the fourth, or N is greater than Mh. Now, if a fourth proportional be found to 999.46458, 1000, and 6282.1056, viz. 6285.461—, then, because 999.46458: 1000 :: 6282.1056: 6285.461and, as before, 999.46458: 1000:: 6282.1056— : N, g 16. 5. h 14. 5. 6282.1056: 6282.1056—:: 6285.461-: N; therefore, since the first of these proportionals is greater than the second, the third is greater than 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. Therefore the circumference has to the diameter a less ratio i 18. 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. It remains to demonstrate that the part by which the circumference exceeds three times the diameter is greater than of the diameter. It has been shown that CG2-750000, therefore CG= 866.02545-, because (866.02545) is greater than 750000; therefore AC+AG=1866.02545-. Now P being, as before, the perpendicular from the centre on the chord of one-twelfth of the circumference, P2 = |