HL is lost than the square of NP; and therefore, since DA is greater than NP, HL is less than NP, and twice HL less than twice NP, wherefore much more is twice HL less than NO. But HL is the difference between half the perimeter of the polygon whose side is EF, and half the circumference of the circle; therefore, twice HL is the difference between the whole perimeter of the polygon and the whole circumference of the circle (5.5.). The difference, therefore, between the perimeter of the polygon and the circumference of the circle is less than the given line NO. COR. 3. Hence, also, a polygon may be inscribed in a circle, such that the excess of the circumference above the perimeter of the polygon may be less than any given line. This is proved like the preceding. PROP. VI. THEOR. The areas of circles are to one another in the duplicate ratio, or as the squares of their diameters. Let ABD and GHL be two circles, of which the diameters are AD and GL; the circle ABD is to the circle GHL as the square of AD to the square of GL. Let ABCDEF and GHKLMN be two equilateral polygons of the same number of sides inscribed in the circles ABD, GHL; and let Q be such a space that the square of AD is to the square of GL as the circle ABD to the space Q. Because the polygons ABCDEF and GHKLMN are equi lateral and of the same number of sides, they are similar (2. 1. Sup., and their areas are as the squares of the diameters of the circles in which they are inscribed. Therefore AD2: GL2 :: polygon ABCDEF : polygon GHKLMN; but AD2: GL2 :: circle ABD : Q; and therefore, ABCDEF : GHKLMN :: circle ABD: Q. Now, circle ABD7ABCDEF; therefore Q7GHKLMN (14. 5.), that is, Q is greater than any polygon inscribed in the circle GHL. In the same manner it is demonstrated, that Q is less than any polygon described about the circle GHL; wherefore the space Q is equal to the circle GHL (2. Cor. 4. 1. Sup.). Now, by hypothesis, the circle ABD is to the space Q as the square of AD to the square of GL; therefore the circle ABD is to the circle GHL as the square of AD to the square of GL. COR. 1. Hence the circumferences of circles are to one another as their diameters. Let the straight line X be equal to half the circumference of the circle ABD, and the straight line Y to half the circumference of the circle GHL X Y And because the rectangles AO.X and GP.Y are equal to the circles ABD and GHL (5. 1. Sup.), therefore AO.X: GP.Y :: AD2 : GL2 :: AO2 ; GP2; and alternately, AO.X: AO2 :: GP.Y: GP2; whence, because rectangles that have equal altitudes are as their bases (1. 6.). X: AO :: Y: GP, and again alternately, X: Y :: AO: GP: wherefore, taking the doubles of each, the circumference ABD is to the circumference GHL as the diameter AD to the diameter GL. COR. 2. The circle that is described upon the side of a right angled triangle opposite to the right angle, is equal to the two circles described on the other two sides. For the circle described upon SR is to the circle described upon RT as the square of SR to the square of RT; and the circle described upon TS is to the circle described upon RT as the square of ST to the square of RT. Wherefore, the circles described on SR and on ST are to the circle described on RT as the squares of SR and of ST to the square of RT (24. 5.). But the squares of RS and of ST are equal to the square of RT (47. 1.); therefore the circles described on RS and ST are equal to the circle described on RT S R T PROP. VII. THEOR. Equiangular parallelograms are to one another as the products of the num bers proportional to their sides. Let AC and DF be two equiangular parallelograms, and let M, N, P and Q be four numbers, such that AB: BC:: M. N; AB: DE: M. and AB: EF:: M: Q, and therefore ex æquali, BC: EF :: N: Q The parallelogram AC is to the parallelogram DF as MN to PQ. P; Let NP be the product of N into P, and the ratio of MN to PQ will be compounded of the ratios (def. 10. 5.) of MN to NP, and NP to PG But the ratio of MN to NP is the same with that of M to P (15. 5.), be cause 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. 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 KL as MM, the square of the number M to NN, the square of the number N. G HK L 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: D::m:s, &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=B2, mxr=nXn, vr=n2. For by this Prop. A.C: B2 :: mxr: n2; but A.C=B2, therefore m×1 =n2. 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:: m : q, then since m, q, r are proportional to A, B, and C, and A.C=B2; therefore, as has just been proved. q2=m Xr: but n2=qxr, by hypothesis, therefore n2=q2, 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 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, sare 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=mx H, 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 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, such that A=360×H+ 375 equal to 360 times H, together with 375 of the 1000 thousandth parts of H. And the same is true in all other cases, where numbers are used to express the relations of geometrical magnitudes. XH, or that A is PROP. VIII. THEOR. The perpendicular drawn from the centre of a circle on the chord of any arc 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 are: And the chord of the arc is a mean proportional between the diameter and a line which is the difference between the radius and the aforesaid perpendicular from the centre Let ADB be a circle, of which the centre is C; DBE any are, 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 circum ference 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 semi circle; and because CGB is also a right angle, the triangles ABD, CBȧ 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 dou ble 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 AD AB.AF (17. 6.), or since AB is 4AH, AD2-4AH.AF. Therefore also, because 4CG2=AD2, 4CG2=4AH.AF, and CG2AH.AF; where fore 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 arc 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 arc DB. 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; 10 the circumference is greater than three times AB, by a line less than 70' In this proposition, the character + placed after a number, signli.es that something is to be added to it; and the character, on the other hand, signifies that something is to be takez wy from it. |