Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

Cor. If two of the nearest sides of an equilateral and equian- Book IV gular figure be bisected, and from the point where these lines cut each other, there be lines drawn to all the angles of the figure, these lines will bifect all the angles of the figure.

PRO P. XIV. PRO B.

To describe a circle about a given equilateral and equiangular

pentagon.

It is required to describe a circle about the equilateral and equiangular pentagon ABCDE.

Bisect the right lines AB, BC^, in the points H and G;a 10. 1. from which draw the right lines HF, GF, intersecting each other in the point F, and at right angles, to AB,BC6; from the b 11. 1. point F draw the right lines BF, FA, FE, FD, FC, they will bifect the angles at A, B, C, D, E. Then, because AB is e-c cor. 13. equal to AE, and AF common, and the angle BAF equal to EAF, the base BF will be equal to EFd. For the same reason, d 4. I. EF is equal to FC; therefore, if, with the center F, and distance B, E, or C, a circle is described, it will pass through the points A, B, C, D, E, of the equilateral and equiangular pentagon : Which was required.

[blocks in formation]

To inscribe an equilateral and equiangular hexagon in a given

circle.

It is required to inscribe the equilateral and equiangular hexagon in the given circle ABCDEF.

Draw AD a diameter to the circle ABCDEF, whose center is G; with the point D as a center, and distance DG, describe a circle EGCH ; join EG, GC ; which produce to the points B, F; join AB, BC, CD, DE, EF, FÅ; then ABCDEF is an equilateral and equiangular hexagon.

For, since G is the center of the circle ABCDEF, GC is e. quala to GD; and since D is the center of the circle CGEH, a def 15.1. GD is equal to DC; therefore CGD is an equilateral triangle b; b 1. 1. but it is likewise equiangular. For the same reason, GDE is c cor. 5. I. an equilateral and equiangular triangle, and equal to the

triangle

Book IV. triangle CGD; and, because BG, GA, are equal to DG, GE,

and the angle BGA to DGE 4, the base AB is equal to DE. d 15. 1. For the same reason, AF is equal to CD; but CD is equal to

DE; therefore AB is equal to AF. Again, because CG falls C 13. 1. upon BE, the angles CGB, CGE, are equal to two right anglese

; but CGD, DGE, are each one third of two right angles f; f cor. 32. 1. therefore CGB is likewise one third of two right angles;

therefore BG, GC, are equal to CG, GF, and the angle BGC equal to the angle CGD; therefore the base BC is equal to CD; but likewise BG, GC, are equal to FG, GE, and the angle BGC to FGE d; therefore the base BC is equal to FE ; but BC is proved equal to CD, and CD to DE; therefore the fix fides BC, CD, DE, EF, FA, AB, are equal; therefore the figure is equilateral; it is likewise equiangular; for the two angles GDC, GDE, are equal to the two angles GCD, GCB; for each is one third of two right anglesf; therefore the whole angle BCD, is equal to the whole CDE. For the same reason, all the other angles are equal to one another ; therefore the figure is likewise equiangular. Wherefore, &c.

Cor. Hence the side of a hexagon is equal to the semi-diameter of the circle. And, if through the points A, B, C, D, E, F, tangents to the circle, be drawn, an equilateral and equiangular hexagon will be described about the circle, as may be proved in the same manner as the pentagon: And so likewise a circle may be inscribed and described about a given hexagon.

PRO P. XVI. PROB.

To infcribe an equilateral and equiangular quindecagon in a

[ocr errors][ocr errors]

It is required to inscribe an equilateral and equiangular quindecagon in the given circle ABCD.

Let AC be the side of an equilateral and equiangular triangle inscribed in the circle, and AB the side of an equilateral and equiangular pentagon, drawn from the point A ; then, if the circle is divided into fifteen parts, the side of the triangle AG will subtend five of them, and the side of the pentagon AB will fubtend three ; therefore BC will be two of said parts; therefore bifect BC in E; BE or CE will be one fifteenth part of the cir.

cumference; if BC, CE, &c. be joined, the equilateral and Book IV. equiangular quindecagon will be inscribed : Which was required.

Cor. If, from what has been said of the pentagon, right lines be drawn through the divisions of the circle, tangents to

an equilateral and equiangular quindecagon will be described about the circle ; or a circle may be inscribed or deg fcribed about the quindecagon.

the same,

THE

[merged small][merged small][ocr errors][merged small][merged small][merged small]

A

1. Book V.

PART is a magnitude of a magnitude; the less of the greater, when the less measures the greater.

II. A multiple is a magnitude of a magnitude; the greater of the less, when the less measures the greater.

III. Ratio is a certain mutual habitude of magnitudes of the fame kind, according to quantity,

IV.
Magnitudes have proportion to each other ; which, being multi-
plied, can exceed one another.

V.
Magnitudes have the same ratio to each other, viz. the first to

the second, and third to the fourth, when there are taken any
equimultiples of the first and third, and likewife any equi-
multiples of the second and fourth; if the multiple of the first
be equal to the multiple of the second, then the multiple of
the third will be equal to the multiple of the fourth; it greatery
greater ; and, if less, lefs.

VI.
Magnitudes which have the same proportion called Propor-

tionals.

[blocks in formation]
[ocr errors]
[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][ocr errors][merged small][ocr errors][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small]
« ΠροηγούμενηΣυνέχεια »