Otherwise.-Find the angular points A, B, C, D, E, of a regular pentagon inscribed in the circle (IV. 11). Through these points draw tangents to the circle ABCDE, and they will form a regular pentagon described about the circle IV. Lemma 2).

To inscribe a circle in a given regular pentagon.

Let ABCDE be the given regular pentagon. It is required to

inscribe a circle in it.

Draw

Bisect the angles BCD and CDE by the straight lines CF and DF (I. 9). From the point F, in which they meet, draw the straight lines FB, FA, and FE. also the perpendiculars FG, FH, FK, FL, and FM to the sides of the pentagon. From centre F, with distance FH, describe the circle GHKLM, B and it is inscribed in the pentagon.

M

C K D

Because the angles FCD and FDC (I. Ax. 7) are equal, F C is equal to F D. Because in the two triangles B C F and DCF, the side B C is equal to the side D C (Hyp.), CF is common, and the angle BCF is equal to the angle DCF (Const.). Therefore B F is equal to D F and also to CF. The angle CB F is also equal to the angle CDF, and is the half of the angle A B C. In the same manner it may be shown that the angles A and E are bisected by the straight lines AF and EF. Because the two angles FCH and F C K are equal (Const.), and the two angles FHC and FK C are also equal, being right angles (Const.). Therefore in the two triangles FHC, FKC, two angles of the one are equal to two angles of the other, and the side F C is common to both. Wherefore, the two triangles are equal (I. 26), and F H is equal to FK. In the same manner, it may be shown that FL, F M, and FG, are each of them equal to FH, or FK. Therefore the five straight lines FG, FH, FK, FL, and FM are equal to one another, and the circle described from the centre F, at the distance of one of them F H, will pass through the extremities of the other four, and touch the straight lines A B, BC, CD, DE, and E A. Because the angles at the points G, H, K, L, and M are right angles, and a straight line drawn through the extremity of the diameter of a circle at right angles to it (III. 16) touches the circle. Therefore each of the straight lines A B, BC, CD. D E, and E A touches the circle. Wherefore the circle GHKLM is inscribed in the pentagon A B C D E. Q. E. F.

Otherwise.--Find F the centre of the inscribed circle (IV. Lemma 1, Cor.), and with radius FH describe the circle G H KLM, and it is inscribed in the pentagon ABCDE.

PROP. XIV. PROBLEM.

To describe a circle about a given regular pentugon.

Let ABCDE be the given regular pentagon. It is required

lescribe a circle about it.

Bisect the angles B CD and CDE by the straight lines CF and F D (I. 9). From the point F, in which they meet, draw the straight lines FB, FA, and FE. With centre F, and distance FC, describe the circle ABCDE, and it is described about the pentagon.

Because it may be shown, as in the preceding proposition, that the side CF is equal to the side FD, and that FB, FA, and FE are each of them equal to FC or FD. Therefore the five straight lines FA, FB, FC, FD, and 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.

Wherefore the circle ABCDE is described about the pentagon ABCDE. Q. E. F. Otherwise.-Find F the centre of the circumscribed circle (IV. Lemma 1, Cor.), and with radius F A describe the circle A B CD E, and it is described about the pentagon ABCDE.

To inscribe a regular hexagon in a given circle.

Let ABCDEF be the given circle. It is required to inscribe a regular hexagon in it.

A

Find the centre G of the circle ABCDEF (III. 1), and draw the diameter A GD. From D, as a centre, at the distance DG, describe the circle EGCH, join EG and CG, and produce them to the points B and F. Join A B, BC, CD, DE, EF, and FA. The hexagon ABCDEF is a regular hexagon.

But

Because G is the centre of the circle ABCDEF, E GE is equal to GD. Because D is the centre of the circle EGCH, DE is equal to DG. Therefore GE is equal to ED (I. Ax. 1), and the triangle EGD is equilateral. Because the three angles EGD, GDE, and DEG, are equal to one another (I. 5, Cor.). the three angles of a triangle are equal to two right angles (I. 32). Therefore the angle E G D is the third part of two right angles. In the same manner it may be shown that the angle D G C is also the third part of two right angles. Because the straight line G C makes with E B the adjacent angles EGC, CGB equal to two right angles (I. 13), the remaining angle CG B is the third part of two right angles. Therefore the angles EGD, DGC, and CGB are equal to one another. And the vertical angles BG A, AGF, and FGE (I. 15) are equal to these angles, each to each. Therefore the six angles EGD, DGC, CG B, BGA, AG F, and FGE are equal to one another. But equal angles stand upon equal arcs (III. 26). Therefore the six arcs A B, BC, CD, DE, EF, and FA are equal to one another. And equal arcs are subtended by equal straight lines (III. 29). Therefore the six straight lines A B, BC, CD, DE, EF, and FA are equal to on another, and the hexagon ABCDEF is equilateral. Again, because the arc AF is equal to the arc ED. To each of these equals, add the arc A B CD. Therefore the whole arc FABCD is equal to the whole arc EDCBA. But the angle FED stands upon the arc FAB CD, and the angle

AFE upon the arc EDCBA. Therefore the angle A FE is equal to the angle FED III. 27). In the same manner it may be shown that the other angles of the hexagon ABCDEF are each equal to the angle AFE or FED. Therefore the hexagon is equiangular. And it was shown to be equilateral. Therefore the regular hexagon ABCDEF s inscribed in the given circle A BCDEF. Q. E. F.

COR. 1.-From this it is manifest, that the side of the hexagon is equal to the straight line drawn from the centre to the circumference, that is, to the radius or semidiameter of the circle.

If through the points A, B, C, D, E, F there be drawn straight lines touching the circle, a regular hexagon will be described about it: and & circle may be inscribed in a given regular hexagon, and circumscribed about it in the same manner as was done in the case of the regular pentagon.

Otherwise.-Find the centre G of the circle ABCDEF, and draw any radius AG. Draw the chord A B equal to A G (IV. 1), and join B G. Because the triangle A G B is equilateral, the angle A G B is one-third of two right angles (1. 32), or one-sixth of four right angles. But all the angles at G are equal to four right angles. Therefore, the arc A B ie one-sixth of the circumference. Draw chords equal to A B, in the circumference all round the circle, and contiguous to each other (IV. 1),—viz., BC, CD, DE, EF, and FA. The figure ABCDEF is a regular hexagon.

Because the chords A B, B C, CD, DE, EF, and FA are all equal, the hexagon is equilateral. Because each of its angles stands upon four-sixths of the circumference, they are equal to each other (III. 27). Therefore the hexagon is equiangular. Wherefore in the circle A B C D E F a regular hexagon is inscribed. Corollary 2.-The interior angles of a regular hexagon are each equal to two-thirds of two right angles, or to a right angle and a third of a right angle. Corollary 3.-A regular hexagon may be described on a given straight line. Corollary 4.-The area of the regular hexagon is six times that of the equilateral triangle described on the same straight line.

Corollary 5.-If the alternate angular points of a regular hexagon be joined, it will form an equilateral triangle. Thus, if A, E, and C be joined an equilateral triangle will be inscribed in the circle; and if tangents to the circle be drawn through these points, an equilateral triangle will be described about it.

PROP. XVI. PROBLEM.

To inscribe a regular quindecagon in a given circle.

Let A B CD be the given circle. It is required to inscribe a regular quindecagon in it.

Find A C the side of an equilateral triangle inscribed in the circle (IV. 2), and A B the side of a regular pentagon inscribed in the same (IV. 11). Bisect B C in E (III. 30). Join B E and E C, and place (IV. 1) in the circumference straight lines equal to these, and contiguous to each other, all round the circle. B The figure A B C D F is a regular quindecagon.

Because of such equal parts as the whole circumference ABCDF contains fifteen, the arc A B C, which is the third part of the whole, contains five; And the arc A B, which is the fifth part of the

E

whole, contains three. Therefore their difference BC contains two of

the same parts, and BE, EC are, each of them, the fifteenth part of the whole circumference A B CD, and the figure A BECDF is equi lateral. Because each of its angles stands upon thirteen-fifteenths of the circumference, it is also equiangular (III. 27). Therefore a regular quindecagon is inscribed in the circle AB C. Q. E. F.

In the same manner as was done in the pentagon, if through the angular points of the inscribed quindecagon, straight lines be drawn touching the circle, a regular quindecagon will be described about it, and likewise, as in the case of the pentagon, a circle may be inscribed in a given regular quindecagon, and circumscribed about it.

Otherwise.-Find two sides of the regular pentagon, and one side of the regular trigon inscribed in the circle. and commencing at the same point in the circumference. Join the points in which the second side of the pentagon and the side of the trigon terminate. The chord thus drawn will be the side of the inscribed pentedecagon or quindecagon. Because of such equal parts as the whole circum. ference contains fifteen, the arc subtended by the two sides of the regular pentagon contains two-fifths or six, and the arc subtended by the side of the trigon contains one-third or five. Therefore their difference contains one fifteenth or one of those parts, which is the side of the quindecagon required.

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.

The term part is here evidently understood in a restricted sense, and one which is commonly expressed by the phrase aliquot part. Better terms are measure

or submultiple, either of which signifies the same as part or aliquot part, and conveys a more distinct notion of the meaning.

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.

The meaning of this definition, taken in connexion with the preceding one, will be best understood by adopting two distinct terms which are correlative; viz., multiple, and submultiple. Thus, if one magnitude contains another an exact number of times, the greater magnitude is called the multiple of the smaller; and the smaller, the submultiple of the greater.

The mutual relation of two agnitudes of the same kind to one another, in respect of quantity, is called their ratio.

The term ratio is employed to express the relation of two like magnitudes to each other, whether they be commensurable or incommensurable, that is, whether they have a common measure or not. Thus, the diagonal of a square has a certain ratio to the side of the square; but this ratio cannot be expressed, like many others, in commensurate terms; for their common measure, or common unit is unknown.

IV.

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

This definition is intended as a test of the likeness or similarity of any two magnitudes; for unless the one can be multiplied so as to exceed the other in magnitude, they cannot be said to be of the same kind, and so cannot have any ratio to each other.

The first of four magnitudes 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