the circle ABC, and AH the half of its circumference. Therefore, &c. Q.E. D.

Book I.

COR. 1. Because DA: AH:: DA2 : DA.AH f, and be- f1.6. caufe by this propofition, DA.AH the area of the circle, of which DA is the radius; therefore, as the radius of any circle to the femicircumference, or as the diameter to the whole circumference, fo is the fquare of the radius to the area of the circle.

COR. 2. Hence, a polygon may be described about a circle, the perimeter of which fhall exceed the circumference of the circle by a line that is less than any given line. Let NO

t

be the given line. Take in NO the part NP less than its half, and lefs also than AD, and let a polygon be defcribed about the circle ABC, fo that its excefs above ABC may be lefs than the fquare of NPs. Let the fide of this polygon be g 1. cor. 4. EF. And fince, as has been proved, the circle is equal to 1. Sup. 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 fquare of NP; and therefore, fince DA is greater than NP, HL is less than NP, and twice HL less than twice NP, wherefore, much more is twice HL lefs than NO. But HL is the difference between half the perimeter of the polygon whofe fide 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 h. The difference, therefore, between the peri- h 5.5. meter of the polygon and the circumference of the circle is lefs than the given line NO.

COR. 3. Hence alfo, a polygon may be infcribed in a circle, fuch that the excefs of the circumference above the perimeter of the polygon may be lefs than any given line. This is proved like the preceding.

PROP.

Supplement

PROP. VI.

THEOR.

THE

HE areas of circles are to one another in the du plicate ratio, or as the fquares, of their dia

meters.

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 fquare of AD to the fquare of GL.

Let ABCDEF and GHKLMN be two equilateral polygons of the fame number of fides infcribed in the circles

ABD, GHL; and let Q be fuch a space that the fquare of AD is to the fquare of GL as the circle ABD to the Ipace Q Because the polygons ABCDEF and GHKLMN are equila a 2. 1. Sup.teral and of the fame number of fides, they are fimilar, and

their

their areas are as the fquares of the diameters of the circles in Book I. which they are inscribed. Therefore, AD2: GL:: poly

gon ABCDEF: polygon GHKLMN; but AD2: GL2:: circle ABD: Q; and therefore, ABCDEF: GHKLM ::

:

circle ABD Q. Now, circle ABD > ABCDEF; therefore Q> GHKLMN b, that is, Qis greater than any poly- b 14. 5. gon infcribed in the circle GHL.

In the fame manner it is demonstrated, that Q is less than any polygon described about the circle GHL; wherefore, the

fpace is equal to the circle GHL c. Now, by hypothefis, c 2. cor. 4 the circle ABD is to the space Q as the fquare of AD to the I. Sup. fquare of GL; therefore the circle ABD is to the circle GHL as the fquare of AD to the fquare of GL. Therefore, &c. Q.E. D.

COR. 1. Hence, the circumferences of circles are to one another as their diameters.

Let the ftraight line X be equal to half the circumference of the circle ABD,

and the ftraight X.

line Y to half the circumference of

the circle GHL: Y

And because the

rectangles AO.X, and GP.Y are equal to the circles ABD

and GHL d; therefore AO.X: GP.Y : : AD2 : GL2 : : AO2 d 5. 1. Sup. :GP2; and alternately, AO.X: AO2 :: GP.Y: GP2; whence, because rectangles that have equal altitudes are as their bafes e, X: AO :: Y: GP, and again alternately, X: e 1. 6. 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.

Supplement COR. 2. The circle that is defcribed upon the fide of a right-angled triangle oppofite to the right angle, is equal to

the two circles defcribed
on the other two fides.
For the circle defcribed up-
on SR is to the circle de-
fcribed upon RT as the
fquare of SR to the fquare.
of RT; and the circle defcri-
bed upon TS is to the circle

defcribed upon KT as the fquare of ST to the fquare of RT. Wherefore, the circles defcribed on SR and on ST are to the circle defcribed on RT as the fquares of SR and of ST to the f 25. 5. fquare of RT f. But the fquares of RS and of ST are equal to 8 47. I. the fquare of RT 8; therefore the circles defcribed on RS and ST are equal to the circle described on RT.

Ε

PROP. VII. THEOR.

QUIANGULAR parallelograms are to one another as the products of the numbers proportional to their fides.

. Let AC and DF be two equiangular parallelograms, and let M, N, P and Q be four numbers, fuch that AB: BC:: M: N; AB: DE:: M: P, and AB: EF:: M: Q, and therefore ex aequali, 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

b 15.5.

NP, and of NP

to PQ. But the

ratio of MN to

C

B D

E

NP is the fame with that of M to Ph, because MN and NP are equimultiples of M and P; and for the fame reason, the

ratio

Book I.

ratio of NP to PQ is the fame 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 fame with that of the fide AB to the fide DE; and the ratio c by Hyp. of N to Q the fame with that of the fide BC to the fide 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 fame ratios d; therefore, the parallelogram AC is to d 23. 6. 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 fquare

defcribed on GH will be to the fquare de

Ꮐ

K

Π

I

fcribed on KL as MM, the fquare of the number M to NN, the fquare 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: D::m:s, &c.; and if the rectangle contained by any two of the lines be equal to the fquare of a third line, the product of the numbers proportional to the first two, will be equal to the fquare of the number proportional to the third; that is, if ̃A.C—B2, mxr=nxn, or = 22.

For by this Prop. A.C: B2: mxr: n2; but A.C=B2, therefore mxrn. Nearly in the fame way it may be demonftrated, that whatever is the relation between the rectangles contained by thefe lines, there is the fame between the products of the numbers proportional to them.

So alfo converfely if m and be numbers proportional to the lines A and C; if alfo A.C=B2, and if a number n be found fuch, that n'mr, then A:B::m: n. For let A: B::m:q, then fince, m, q, r are proportional to A, B, and C, and A.C=B2; therefore, as has just been proved, qmxr; but nqxr, by hypothefis, therefore n22, and n wherefore A: B::m: n.

SCHOLIUM.