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about the circle A, but what must exceed it by a space greater than C, which is absurd. In the same manner, if B be less than A by the space C, it is shewn that no polygon can be inscribed in the circle A, but what is less than A by a space greater than C, which is also absurd. Therefore, A and B are not unequal; that is, they are equal to one another.

PROP. V. THEOR.

The area of any circle is equal to the rectangle contained by the semi-diameter, and a straight line equal to half the circumference.

Let ABC be a circle of which the centre is D, and the diameter AC; if in AC produced there be taken AH equal to half the circumference, the area of the circle is equal to the rectangle contained by DA and AH.

Let AB be the side of any equilateral polygon inscribed in the circle ABC; bisect the circumference AB in G, and through G draw EGF touching the circle, and meeting DA produced in E, and DB produced in

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F; EF will be the side of an equilateral polygon described about the circle ABC (3. 1. Sup.). In AC produced take AK equal to half the perimeter of the polygon whose side is AB; and AL equal to half the perimeter of the polygon whose side is EF. Then AK will be less, and AL greater than the straight line AH (Lem. Sup.). Now, because in the triangle EDF, DG is drawn perpendicular to the base, the triangle EDF

is equal to the rectangle contained by DG and the half of EF (41. 1.); and as the same is true of all the other equal triangles having their vertices in D, which make up the polygon described about the circle; therefore, the whole polygon is equal to the rectangle contained by DG and AL, half the perimeter of the polygon (1. 2.), or by DA and AL. But AL is greater than AH, therefore the rectangle DA.AL is greater than the rectangle DA.AH; the rectangle DA.AH is therefore less than the rectangle DA.AL, that is, than any polygon described about the circle ABC.

Again, the triangle ADB is equal to the rectangle contained by DM the perpendicular, and one half of the base AB, and it is therefore less than the And as the same rectangle contained by DG, or DA, and the half of AB

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is true of all the other triangles having their vertices in D, which make up the inscribed polygon, therefore the whole of the inscribed polygon is less than the rectangle contained by DA, and AK half the perimeter of the polygon. Now, the rectangle DA.AK is less than DA.AH; much more, therefore, is the polygon whose side is AB less than DA.AH; and the rectangle DA.AH is therefore greater than any polygon inscribed in the circle ABC. But the same rectangle DA.AH has been proved to be less than any polygon described about the circle ABC; therefore the rectangle DA.AH is equal to the circle ABC (2. Cor. 4. 1. Sup.). Now DA is the semidiameter of the circle ABC, and AH the half of its circumference.

COR. 1. Because DA: AH:: DA2: DA.AH (1. 6.), and because by this proposition, DA.AH= the area of the circle, of which DA is the radius: therefore, as the radius of any circle to the semicircumference, or as the diameter to the whole circumference, so is the square of the radius to the area of the circle.

COR. 2. Hence a polygon may be described about a circle, the perimeter of which shall exceed the circumference of the circle by a line that is less than any given line. Let NO be the given line. Take in NO the part NP less than its half, and also than AD, and let a polygon be described about the circle ABC, so that its excess above ABC may be less than the square of NP (1. Cor. 4. 1. Sup.). Let the side of this polygon be EF. And since, as has been proved, the circle is equal to the rectangle DA.AH, and the polygon to the rectangle DA.AL, the excess of the polygon above the circle is equal to the rectangle DA.HL; therefore the rectangle DA.

HL is less 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

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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 equilateral 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.

Con. 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 ᎡᎢ. .

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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:

P; 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.

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 PQ. But the ratio of MN to NP is the same with that of M to P (15. 5.), be

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

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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 mxr 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

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