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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 (6, Cor. 2). Now, DA is the semidiameter of the circle ABC, and AH the half of its circumference.

COR. 1.-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 less 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 (6, Cor. 1). 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 twice HL is still 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. The difference, therefore, between the perimeter of the polygon and the circumference of the circle, is less than the given line NO. COR. 2.-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. COR. 3.-Hence (VI. 33), the area of a sector is equal to half the rectangle under its arc and the radius.

PROPOSITION VIII. THEOREM.

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 diame

ters 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 A

that the square of AD is to

B

the square of F

E

G

H

K

M

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 (4), and are as the squares of the diameters of the circles in which they are inscribed; therefore, as the square of AD to the square of GL, so is the polygon ABCDEF to the polygon GHKLMN; but as the square of AD to the square of GL, so is the circle ABD to the space Q; therefore the polygon ABCDEF is to the polygon GHKLMN, as the circle ABD to the space Q; but the polygon ABCDEF is less than the circle ABD, therefore GHKLMN is less than the space Q (V. 14); wherefore the space Q is greater than any polygon inscribed in the circle GHIL.

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 (4, Cor. 2). 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

and because the rectangles AO X and CP Y are equal to the circles Y

ABD and GHL (7); therefore the rectangle AO X is to the rectangle GP Y, as the square of AD to the of GL, or as the square of AO to the square of GP; there

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fore, alternately the rectangle AO X is to the square of AO, as the rectangle GP Y to the square of GP; but rectangles that have equal altitudes are as their bases, therefore X is to AO as Y to GP; and again, alternately, X is to Y as AO to GP, and taking the doubles of each, the circumference ABD is to the circumference GHL as the diameter AD to the diameter GL.

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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 RT as the upon square of R 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 ST are to the circle described on RT as the squares of SR and ST to the square of RT (V. 24); but the squares of RS and ST are equal to the square of RT; therefore the circles described on RS and ST are equal to the circle described on RT.

PROPOSITION IX. THEOREM.

Equiangular parallelograms are to one another as the products of the numbers 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, by equality, 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 of MN to NP, and of NP to PQ; but the ratio of MN to NP is the same with that of M

to P, because MN and NP are equimultiples of M and P; and A

B D

C

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E

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. ratio of M to P is the same with that of the side

Now, the AB to the

side DE; 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; 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.

G

H K

L

COR. 1.-Hence, if GH be to KL as the number M to 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. COR. 2.-If the square on GH be to the square on KL as a number to a number, the line GH will be to the line KL as the square root of the first number to the square root of the second number; for, if GH were to KL in any other ratio, the square described on GH would not be to the square described on KL in the ratio supposed.

Schol. This proposition is demonstrated here only in the case of commensurable lines, as the fifteenth proposition of the fifth book, on which part of the demonstration depends, is proved only in the case of integral multiples. Before it be proved for incommensurable lines, it must first be proved that M: PMN: NP when N is an interminate number (Ad. V. Def. 3, and Al. Pr. VIII.*) The numbers proportional to the sides are the numbers denoting the number of times that the unit of measure is contained in these sides (Ad. V. Def. 5).

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The area of a rectangle is equal to the square of the unit of measure multiplied by the product of the numbers denoting the number of times that this unit is contained in two of its adjacent sides.

Let CDEF be a rectangle, M the unit of measure, and let its sides CD, DE, be denoted by A and B, and the num

* The reference here is to the additional fifth book and the algebraical principles at the end of the volume.

ber of times that M is contained in A and B respectively by a and b, then A⚫BabM2.

F

E

1. When M is contained a certain number of times in A and B, as 4 and 3 times respectively, it is obvious that CE may be divided into a number of squares each = M2, by drawing lines parallel to the sides through the points of division, that there will be as many rows as there are units in b, and as many squares in each row as there are units in a, therefore generally M A BabM2.*

C

2. When a and b are terminate, but not integral num

m

bers, let a = and b =2, these are equal to and

n

пр

mq
nq' ng

Take a line M' such that M= nqM', then A and B will contain M' as often as there are units in mq and np. Call these products a' and b' respectively, then (by 1st case) A B = a'b'M'2 = mqnpM'2. But M2nqnq M'2 (also by

1st case)

M2

neq2M22, or M22= ;
n2q2

mnpqM2 mp M2. = m. PM2.

n2q2

=

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therefore AB=

3. Let b be an interminate number. If AB is not = abM2, let A B = ab'M2; and let b' be interminate and b, and let b′′ be an intermediate terminate number. Then (by 2d case) A· B" ab"M2, where B" is such a line that B: B′′ =b: b′′; and since bb′′, therefore B <B′′ (V. 14), and A BAB", and therefore ab'M2 ab"M2. But b'b", therefore ab'M2 is also > ab"M2, which is impossible; therefore b' is not b; and it may be similarly shown that it is not less; therefore A BabM2.·

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4. Let a and b be both interminate. Then if AB is not = abM2, let AB ab'M2, b' being b as in the 3d case, and let B" and b′′ be taken as in that case, then A⚫B"

* Before the student can understand any of the following cases, it will be necessary for him to be acquainted with the algebraical principles at the end.

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