Sidebilder
PDF
ePub

angle FLC: and because KC is equal to CL, KL is double of KC: in the same manner, it may be shown that HK is double of BK: and because BK is equal to KC, as was demonstrated, and that KL is double of KC, and HK double of BK, HK shall be equal to KL: in like manner it may be shown that GH, GM, ML are each of them equal to HK or KL: therefore the pentagon GHKLM is equilateral. It is also equiangular; for, since the angle FKC is equal to the angle FLC, and that the angle HKL is double of the angle FKC, and KLM double of FLC, as was before demonstrated, the angle HKL is equal to KLM: and in like manner it may be shown, that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM: therefore the five angles GHK, HKL, KLM, LMG, MGH being equal to one another, the pentagon GHKLM is equiangular : and it is equilateral, as was demonstrated; and it is described about the circle ABCDE. Which was to be done.

PROP. XIII. PROB.

To inscribe a circle in a given equilateral and equiangular pentagon.

[ocr errors]

Let ABCDE be the given equilateral and equiangular pentagon: it is required to inscribe a circle in the pentagon ABCDE.

Bisect (9. 1.) the angles BCD, CDE by the straight lines CF, DF, and from the point F, in which they meet, draw the straight lines FB, FA, FE; therefore, since BC is equal to CD, and CF common to the triangles BCF, DCF, the two sides BC, CF, are equal to the two DC, CF; and the angle BCF is equal to the angle DCF; therefore the base BF is equal (4. 1) to the base FD, and the other angles to the other angles, to which the equal sides are opposite; therefore the angle CBF is equal to the angle CDF: and because the angle CDE is double of CDF, and that CDE is equal to CBA, and CDF to CBF; CBA is also double of the angle

A CBF; therefore the angle ABF is equal

G to the angle CBF; wherefore the angle

M ABC is bisected by the straight line BF: in the same manner it may be demon

B strated, that the angles BAE, AED are

TE bisected by the straight lines AF, FE: from the point F draw (12. 1.) FG, FH,

Н. FK, FL, FM, perpendiculars to the

L straight lines AB, BC, CD, DE, EA; and because the angle HCF is equal to KCF, and the right angle FHC equal to

C K D the right angle FKC; in the triangles FHC, FKC there are two angles of one equal to two angles of the other, and the side FC, which is opposite to one of the equal angles in each, is common to both; therefore the other sides shall be equal (26. 1.) each to each ; wherefore the perpendicular FH is equal to the perpendicular FK: in the same manner it may be demonstrated

that FL, FM, FG are each of them equal to FH, or FK; therefore the five straight lines FG, FH, FK, FL, FM are equal to one another: wherefore the circle described from the centre F, at the distance of one of these five, shall pass through the extremities of the other four, and touch the straight lines AB, BC, CD, DE, EA, because the angles at the points G, H, K, L, M are right angles; and that a straight line drawn from the extremity of the diameter of a circle at right angles to it, touches (16. 3.) the circle: therefore each of the straight lines AB, BC, CD, DE, EA touches the circle; wherefore it is inscribed in the pentagon ABCDE. Which was to be done.

PROP. XIV. PROB.

F

To describe a circle about a given equilateral and equiangular pentagon.

Let ABCDE be the given equilateral and equiangular pentagon; it is required to describe a circle about it.

Bisect (9. 1.) the angles BCD, CDE by the straight lines CF, FD, and from the point F, in which they meet, draw the straight lines FB, FA, FE, to the points B, A, E. It

A may be demonstrated, in the same manner as in the preceding proposition, that the angles CBA, BAE, AED, are bisected by the B straight lines FB, FA, FE: and because the

E angle BCD is equal to the angle CDE, and that FCD is the half of the angle BCD, and CDF the half of CDE; the angle FCD is equal to FDC: wherefore the side CF is equal (6. 1.) to the side FD : in like manner

С

D it may be demonstrated that FB, FA, FE are each of them equal to FC or FD: therefore the five straight lines FA, FB, FC, FD, FE are equal to one another; and the circle described from the centre F, at the distance of one of them, shall pass (through the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE. Which was to be done.

PROP. XV. PROB. To inscribe an equilateral and equiangular hexagon in a given circle.*

Let ABCDEF be the given circle; it is required to inscribe an equilateral and equiangular hexagon in it.

Find the centre G of the circle ABCDEF, and draw the diameter AGD; and from D as a centre, at the distance DG, describe the circle EGCH, join EG, CG, and produce them to the

* See Note.

points B, F; and join AB, BC, CD, DE, EF, FA: the hexagon ABCDEF, is equilateral and equiangular.

Because G is the centre of the circle ABCDEF, GE is equal to GD: and because D is the centre of the circle EGCH, DE is equal to DG: wherefore GE is equal to ED, and the triangle EGD is equilateral; and therefore its three angles EGD, GDE, DEG are equal to one another, because the angles at the base of an isosceles triangle are equal (5. 1.); and the three angles of a triangle are equal (32. 1.) to two right angles; therefore the angle EGD is the third part of two right angles : in the same manner, it may be demonstrated, that the angle DGC is also the third

A part of two right angles: and because the straight line GC makes with EB the adja

F cent angles EGC, CGB equal (13. 1.) to two G

B right angles; the remaining angle CGB is the third part of two right angles; therefore the angles EDG, DGC, CGB are equal to one another : and to these are equal (15. 1.) E the vertical opposite angles BGA, AGF, FGE: therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE are equal to one another: but equal angles stand upon equal

D (26. 3.) circumferences; therefore the six circumferences AB, BC, CD, DE, EF, FA are equal to one another: and equal circum

H ferences are subtended by equal (29. 3.) straight lines ; therefore the six straight lines are equal to one another, and the hexagon ABCDEF, is equilateral. It is also equiangular; for, since the circumference AE is equal to ED, to each of these add the circumference ABCD: therefore the whole circumference FABCD shall be equal to the whole EDCBA: and the angle FED stands upon the circumference FABCD, and the angle AFE upon EDCBA; therefore the angle AFE is equal to FED : in the same manner it may be demonstrated that the other angles of the hexagon ABCDEF are each of then equal to the angle AFE or FED; therefore the hexagon is equiangular; and it is equilateral, as was shown; and it is inscribed in the given circle ABCDEF. Which was to be done.

Cor. From this it is manifest, that the side of the hexagon is equal to the straight line from the centre, that is, to the semidiameter of the circle.

And if through the points A, B, C, D, E, F there be drawn straight lines touching the circle, an equilateral and equiangular hexagon shall be described about it, which may be demonstrated from what has been said of the pentagon; and likewise a circle may be inscribed in a given equilateral and equiangular hexagon, and circumscribed about it, by a method like to that used for the pentagon.

PROP. XVI. PROB.

To inscribe an equilateral and equiangular quindecagon, in a given circle. *

Let ABCD be the given circle; it is required to inscribe an equilateral and equiangular quindecagon in the circle ABCD.

Let AC be the side of an equilateral triangle inscribed (2. 4.) in the circle, and AB the side of an equilateral and equiangular pentagon inscribed (11. 4.) in the same; therefore, if such equal parts as the whole circumference ABCDF contains fifteen, the circumference ABC, being the third part of the whole, contains 'five; and the circumference AB, which is the fifth part of the whole, contains three; therefore BC their difference B

F contains two of the same parts : bisect (11. 4.) BC in E; therefore BE, EC are, each of them, the fifteenth part of the E whole circumference ABCD: therefore, if the straight lines BE, EC be drawn, and

D straight lines equal to them be placed (1. 4.) around in the whole circle, an equilateral and equiangular quindecagon shall be inscribed in it. Which was to be done.

And in the same manner as was done in the pentagon, if through the points of division made by inscribing the quindecagon, straight lines be drawn touching the circle, an equilateral and equiangular quindecagon shall be described about it: and likewise as in the pentagon, a circle may be inscribed in a given equilateral and equiangular quindecagon, and circumscribed about it.

* See Note.

THE

ELEMENTS OF EUCLID.

BOOK V.

DEFINITIONS.

I.

A LESS magnitude is said to be a part of a greater magnitude, when

the less measures the greater, that is, when the less is contained a certain number of times exactly in the greater.'

II. A greater magnitude is said to be a multiple of a less, when the

greater is measured by the less, that is, when the greater contains the less a certain number of times exactly.'

*Ratio is a mutual relation of two magnitudes of the same kind to one another, in respect of quantity.' *

IV. Magnitudes are said to have a ratio to one another, when the less can be multiplied so as to exceed the other.

V. The first of four maguitudes, is said to have the same ratio to the

second, which the third has to the fourth, when any equimultiples whatsoever of the first and third being taken, and any equimultiples whatsoever of the second and fourth; if the multiple of the first be less than that of the second, the multiple of the third is also less than that of the fourth; or, if the multiple of the first be equal to that of the second, the multiple of the third is also equal to that of the fourth; or, if the multiple of the first be greater than that of the second, the multiple of the third is also greater than that of the fourth.

VI. Magnitudes which have the same ratio are called proportionals.

* See Note.

« ForrigeFortsett »