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Suppl. greater than the straight line AHb. Because DG is drawn
perpendicular to the base, the triangle EDF is equal to the b Ax. 1.
rectangle contained by DG and the half of EF¢; and the Sup.
same is true of all the other equal triangles having their verc 41. 1.
tices in D, which make up the polygon described about the
circle ; therefore the whole polygon is equal to the rectangle d 1. 2.
contained by DG and AL, half the perimeter of the polygona, or by DA and AL. But AL is greater than AH; therefore the rectangle DA.AL is greater than the rectangle DA.AH, or the rectangle DA.AH is 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 the perpendicular DM, and half of the base AB, and is therefore less than the rectangle contained by DG, or DA, and half of AB. The same is true of all the other triangles having their vertices in D, which make up the inscribed polygon. Therefore the whole 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; therefore the rectangle DA.AH is greater than any polygon inscribed in the circle ABC. But the 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 Now DA is 4. 1. Sup. the semidiameter of the circle ABC, and AH half of its cir
cumference. Therefore, the area of any circle, &c. Q. E. D.
e 2. Cor.
f 1. 6.
CoR. Because DA : AH :: DAP: DA.AHf, and because DA.AH=the area of the circle, of which DA is the radius, as the radius of any circle is to the semicircumference, or as the diameter is to the whole circumference, so is the square of the radius to the area of the circle,
PROP. VI. PROB.
TO describe a polygon about a circle, the perimeter of which shall exceed the circumference of the circle by a line that is less than any given line. See the last figure.
Let ABC be the given circle, and NO the given line; it is Book I. required to describe about the circle ABC a polygon of which the perimeter shall exceed the circumference of the circle by a line that is less than NO.
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 NP2. Let the side of this polygon be EF. It may be shown, a 1. Cor.
4. I. Sup; as in the last proposition, that the circle is equal to the rectangle DA.AH, and the polygon to the rectangle DA.AL; therefore 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. But DA is greater than NP, therefore 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 circleb. Therefore the difference between the perime- b 5.5. ter of the polygon and the circumference of the circle is less than the given line NO. Therefore, a polygon has been described about a circle, &c. Which was to be done.
CoR. 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. VII. THEOR.
THE areas of circles are to one another in the du. plicate 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 and 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 a 2. 1. Sup. equilateral and of the same number of sides, they are similara,
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 AD: GL :: circle ABD :Q; therefore ABCDEF : GHKLMN::
b 14. 5.
circle ABD: Q. Now the circle ABD>ABCDEF, therefore Q>GHKLMN, that is, Q is greater than any polygon inscribed in the circle GHL. · In the same manner it may
be demonstrated that Q is less than any polygon described about the circle GHL; wherefore the space Q is equal to the circle GHL. Now, by hypothesis, the circle ABD : space Q:: AD2 : GL? ; therefore the circle ABD: circle GHL :: ADP : GL%. Therefore, circles, &c. Q. E. D.
Cor. The circle 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 upon SR: circle upon RT :: SR : RT2, and the circle upon ST: circle upon RT :: ST2 : RT2 ;
THE circumferences of circles are to one another as their diameters. See the figure, Prop. VII.
Let the straight line X be equal to half the circumference of the circle ABD, and the straight
X line Y to half the circumference of
Ythe circle GHL. Because the rectangles AO.X and GP.Y are equals to the circles ABD and a 5. 1. Sup. GHL, A0.X : GP.Y :: AD2: GL' :: A0: GP2; there- b 7.1. Sup. fore, alternatelye, A0.X : A0% :: GP.Y : GP2; therefore c 16.5. X: A0 :: Y: GPd; therefore X: Y:: AO: GP®; where- d B. 5. fore the circumference ABD: circumference GHL : : diameter AD: diameter GL. Therefore, the circumferences, e 15. 5. &c. Q. E. D.
PROP. IX. THEOR.
EQUIANGULAR parallelograms are to one ano. ther as the products of the numbers proportional to their sides,
Suppl. Let AC and DF be two equiangular parallelograms, and
let M, N, P, Q be four numbers such that AB : BC:;M:
N, AB : DE :: M:P, and AB : EF::M:Q, and therea 22. 5. fore, ex æqualia, BC: EF::N:Q; the parallelogram AC
: parallelogram DF:: MN: PQ.
Let NP be the product of N into P, and the ratio of MN to PQ will be
C compounded of bDef, 10.5. the ratiost of
F MN to NP, and of NP to PQ, But the ratio of MN to
NP is the same c 15. 5. with that of M to Ps, because MN and NP are equimulti
ples of M and P; and 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. 'But the
ratio of M to P is the same with that of the side AB to the d by Hyp. side Ded; and the ratio of N to Q is 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. But the ratio of the parallelogram AC to the parallele 23. 6.
ogram DF is compounded of the same ratiose; 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. Therefore, equiangular parallelograms, &c. Q. E, D,
Cor. 1. Hence, if GH be to KL as the number M to the number N, the square described on GH will
H K L be to the square describ
as MM, the square of the number M, to NN, the square of the number N.
ed on KL,
COR. 2. If A, B, C, D, &c. be any lines, and m, n, r, s, &c. numbers proportional to them, 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 lines will be equal to the square of the number proportional to the third line; that is, if A.C=B2, then mxr=nxn=n?.