equal to KC, as was demonstrated, and KL is double of KC, and HK double of BK, HK is equal to KL; in like manner, it may be shewn 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 the angle HKL 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 shewn, 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 demonstra ted and it is described about the circle ABCDE.

PROP. XIII. PROB.

To inscribe a circle in a given equilateral and equiangu.ar pentagon. Let ABCDE be the given equilateral and equiangular pentagon; it is required to inscribe a circle in the pentagon ABCDE.

G

H

K

M

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 CDE equal to CBA, and CDF to CBF; CBA is also double of the angle CBF; therefore the angle ABF is equal to the angle CBF; wherefore the angle ABC is bisected by the straight line BF: in the same manner, it may be demonstrated that the angles BAE, AED, are bisected by the straight lines AF, EF: B from the point F draw (12. 1.) FG, FH, FK, FL, FM perpendiculars to the straight lines AB, BC, CD, DE, EA; and because the angle HCF is equal to KCF, and the right angle FHC equal to 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, will pass through the extremities of the other four, and touch the straight lines AB, BC, CD, DE, EA, because that 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 (1. Cor. 16. 3.) the circle; therefore each of the straight lines AB, BC, CD, DE, EA touches the circle; wherefore the circle is inscribed in the pentagon ABCDE.

PROP. XIV. PROB.

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.

A

E

F

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 may be demonstrated, in the same manner as in the preceding proposition, that the angles CBA, BAE, AED are bisected by the straight lines FB, FA, FE: and because that the 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 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, will pass through the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE.

PROP. XV. PROB.

D

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 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 (Cor. 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 part of two right angles; and because the straight line GC makes with EB the adjacent angles EGC, CGB equal (13. 1.) to two right angles; the remaining angle CGB is the third part of two right angles; therefore the angles EGD, DGC, CGB, are equal to one another; and also the angles vertical to them, BGA, AĞF, FGE (15.

F

A

B

1.); therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE are equal to one another. But equal angles at the centre stand upon equal arcs (26. 3.): therefore the six arcs AB, BC, CD, DE, EF, FA are equal to one another and equal arcs 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 arc AF is equal to ED, to each of these add the arc ABCD; therefore the whole arc FABCD shall be equal to the whole EDCBA: and the angle FED stands upon the arc 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 them equal to the angle AFE or FED; therefore the hexagon is equiangular; it is also equilateral, as was shown; and it is inscribed in the given circle ABCDEF. 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 radius of the circle.

H

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.

B

F

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, of such equal parts as the whole circumference ABCDF contains fifteen, the arc ABC, being the third part of the whole, contains five; and the arc AB, which is the fifth part of the whole, contains three; therefore BC their differ-E ence contains two of the same parts: bisect (30. 3.) BC in E; therefore BE, EC are, each of them, the fifteenth part of the 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 will be inscribed in it.

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 may 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.

SCHOLIUM.

Any regular polygon being inscribed, if the arcs subtended by its sides be severally bisected, the chords of those semi-arcs will form a new regular polygon of double the number of sides: thus, from having an inscribed square, we may inscribe in succession polygons of 8, 16, 32, 64, &c. sides; from the hexagon may be formed polygons of 12, 24, 48, 96, &c. sides; from the decagon polygons of 20, 40, 80, &c. sides; and from the pentedecagon we may inscribe polygons of 30, 60, &c. sides; and it is plain that each polygon will exceed the preceding in surface or area.

It is obvious that any regular polygon whatever might be inscribed in a circle, provided that its circumference could be divided into any proposed number of equal parts; but such division of the circumference like the trisection of an angle, which indeed depends on it, is a problem which has not yet been effected. There are no means of inscribing in a circle a regular heptagon, or which is the same thing, the circumference of a circle cannot be divided into seven equal parts, by any method hitherto discovered. It was long supposed, that besides the polygons above mentioned, no other could be inscribed by the operations of elementary Geometry, or, what amounts to the same thing, by the resolution of equations of the first and second degree. But M. Gauss, of Göttingen, at length proved, in a work entitled Disquisitiones Arithmetica, Lipsie, 1801, that the circumference of a circle could be divided into any number of equal parts, capable of being expressed by the formula 2"+1, provided it be a prime number, that is, a number that cannot be resolved into factors.

The number 3 is the simplest of this kind, it being the value of the above formula when n=1; the next prime number is 5, and this is also contained in the formula; that is, when n= =2. But polygons of 3 and 5 sides have already been inscribed. The next prime number expressed by the formula is 17; so that it is possible to inscribe a regular polygon of 17 sides in a circle.

For the investigation of Gauss's theorem, which depends upon the theory of algebraical equations, the student may consult Barlow's Theory of Numbers.

14

ELEMENTS

OF

GEOMETRY.

BOOK V.

IN the demonstrations of this book there are certain" signs or characters" which it has been found convenient to employ.

1. The letters A, B, C, &c. are used to denote magnitudes of any kind. The letters m, n, p, q, are used to denote numbers only.

It is to be observed, that in speaking of the magnitudes A, B, C, &c., we mean, in reality, those which these letters are employed to represent; they may be either lines, surfaces, or solids.

"2. When a number, or a letter denoting a number, is written close to "another letter denoting a magnitude of any kind, it signifies that the "magnitude is multiplied by the number. Thus, 3A signifies three "times A; mB, m times B, or a multiple of B by m. When the num"ber is intended to multiply two or more magnitudes that follow, it is "written thus, m(A+B), which signifies the sum of A and B taken m "times; m(A-B) is m times the excess of A above B.

"Also, when two letters that denote numbers are written close to one an"other, they denote the product of those numbers, when multiplied into one another. Thus, mn is the product of m into n; and mnA is A mul"tiplied by the product of m into n.

66

DEFINITIONS.

1. 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.

2. 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.

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