the arc, whose centre is A, in E: and produce BG to meet ID in D. Then join CD, DE, EA; and ABCDE is the pentagon required.

For, joining AC, AD, BE, each of the angles AEB, ADB, ACB may easily be shewn to be two-fifths of a right angle; .. these angles being on the same base AB, and equal to one another, they are angles in the same segment', that is, A, B, C, D, E are points in the circumference of the same circle. Supposing this circle drawn, since EBD=two-fifths of a right angle = < CBD = 4 ADB, and since equal angles, whether at the centre or circumference, in the same circle are subtended by equal chords, .. ED=CD= AB; and BC= AB = AE;

ABCDE is equilateral. And it is also equiangular, for ABC was made equal to six-fifths of a right angle (86 Cor. 1), and each of the other angles will easily be shewn to be equal to this.

164. PROP. LXVI. To construct a regular hexagon with each of its sides equal to a given line or length.

Let AB be the given line or length; produce it both ways to a, and b, making Aa = Bb

AB. Upon ab describe an equilateral triangle (23), on that side of it on which the hexagon is to lie; and divide each of the sides bc, ac, into three equal parts in the points C, D, and E, F. Join AF, BC, DE, and ABCDEF shall be the hexagon required.

F

E

a

A

B

D

For the triangles AFa, BCb, DEc being equilateral and equiangular, and equal to one another in all respects, the proof is obvious.

This construction enables us to lay down a regular hexagon on the ground with remarkable ease and certainty. For it may be done evidently with any long staff, or tape, or chord, only, on which the given length of the side is marked.

165. PROP. LXVII. To construct a regular octagon with each of its sides equal to a given line or length.

Let AB be the given line or length; produce it in

definitely to b; from A and B

draw AF, BE at right angles to AB, on that side of it on which the octagon is required. Bisect

EL

D

A B

EBb by the straight line BC, making BC equal to AB. (If the compasses from the first be set to AB, they will need no change through the whole operation; for the angle EBb may be bisected with that radius as well as any other.) Through C draw CD parallel to BE, and equal to AB; with centre D describe an arc cutting BE in E; through E draw EF parallel to AB; through F draw FG parallel to BC, and equal to AB; through G draw GH parallel to AF, and equal to AB; and join AH. ABCDEFGH is the octagon required.

For all the sides, except AH, are made equal to one another; and by joining HF, and BD, the triangles AFH, BED may easily be shewn to be equal in all respects (24); and .. AH=DE=AB. :. ABCDEFGH, which has eight sides, is equilateral. It is also equi-angular; for 4 ABC, by construction, is equal to one right angle and a half. Also, ▲ DCc = 4 EBC (34) = half a right angle, .. ‹ BCD = one right angle and a half (30) = ABC. And the same may be proved of the other angles. (See 86 Cor. 1).

166. PROP. LXVIII. To cut the greatest regular octagon out of a given square.

D

Let ABCD be the given square. AC, BD intersecting in E. With centre A and radius AE, mark off AF in AB; and AG in AD; and do the same from each of the other corners, so that AE = AF = AG=BH=BI=CK=CL=DM = DN. Join FL, HN, KG, MI, and IFLHNKGM shall be the octagon required.

For, since the sides of the square are equal, it is obvious

PART II.

A

Draw its diagonals

.'.

=

enough from the construction, that IF=HL= KN = GM. But it is not so obvious that the alternate sides are equal to these and to one another, for instance that GK = KN. To prove this, join EG, EK, EN. Then, since AG=AE, LAGE= = ▲ AEG (26); and ▲ EAG=half a right angle, LAGE + AEG one right angle and a half (37), and AGE= three-fourths of a right angle. But DG = DK, .. < DGK = half a right angle, .. 4 EGK= three-fourths of a right angle. In the same manner it may be shewn, that 4 EKG = EKN three-fourths of a right angle = ENK; .. the two triangles EGK, EKN have two angles in one equal to two angles in the other, each to each, and one side common, .. the triangles are equal, and the sides are equal which are opposite to equal angles (39), that is, GK KN. The same may be proved for any other two adjacent sides of the octagon; .. it is equilateral. It is also equi-angular, since each of its angles is obviously equal to one right angle and a half.

=

=

167. OBS. Constructions have now been given for regular figures of 3, 4, 5, 6, 8, and 10, sides. Of course, by bisecting the arcs of the circumscribing circle subtended by these sides in any case, a regular polygon of double the number of sides is obtained, that is, we can add to the above other polygons of 12, 16, and 20 sides; and again, by subdividing, we have polygons of 24, 32, and 40 sides; and so on. It is to be observed, that, while we do this, we at the same time divide the circumference of a circle into as many equal parts as the polygon has sides; and thus the whole circumference of any circle may be minutely and equally subdivided in a great variety of ways. But it is not true, that we can in this way divide the circumference of a circle into any proposed number of equal parts, because in attempting this we shall often find ourselves brought to the necessity of trisecting an arc, a problem for which no strictly geometrical solution has yet been discovered. We can trisect some particular arcs, as the arc of a quadrant, of a semicircle, and of a whole circle, but not of any arc; and so we are hindered from effecting with theoretical exactness that minute subdivision of the circle which is in common use for scientific purposes. The consequence is that, for most practical purposes, both whole circumferences of

given circles, and given arcs are divided into equal parts by trial, either with the ordinary compasses alone, as in (169), or with the help of another circle which is itself already graduated, called a Protractor. Such an instrument, and its use, will be explained in Part III.

PROPORTIONAL LINES AND AREAS.

168. PROP. LXIX. To divide a given straight line into any proposed number of equal parts.

(1) This may be done, in most cases, by trial with the compasses alone without any appreciable error. Thus let AB be the given line, which is required to be divided, suppose, into 5 equal parts. Make a guess at the 5th part

A

A

C B

B C

of AB; take this distance in the compasses, and step along AB from A to B. If in the 5th step the foot of the compasses falls exactly upon B, then, by marking each step on the line, the thing required is done.-But it is more probable, that the given line will not be thus equally subdivided at the first trial. The 5th step of the compasses will be more likely either to fall short, or pass beyond, the point B, by the short length BC. Then, if this short length be divided into 5 equal parts by the eye, (which may be done with sufficient accuracy for most practical purposes), and one of such parts be added to the distance in the compasses, or subtracted from it, as the case may be, then with the distance so corrected step along AB, and it will divide AB into the required number of equal parts.

(2) The theoretically exact method of doing the same thing is given in (68); and another method of a similar kind is as follows:-Let AB be the the given straight line which it is required to divide into 5 equal parts. From A draw AI making any acute angle with AB; produce IA to any point C; through C draw an indefinite straight line parallel to AB; and taking a

distance in the compasses, at a guess, somewhat greater than the 5th part of AB, step along this line, beginning from C, marking the points of division so that CD= DE=EF=FG=GH.

Join HB, and produce

it to meet CI in the point I. Then join ID, IE, IF, IG, intersecting AB in the points d, e, f, g; and AB shall be divided by these points as required.

For, since ICD, IAd are similar triangles, by (71) Ad CD: Id: ID. Similarly de: DE: Id: ID, Ad: CD: de: DE; and, alternately, Ad: de :: CD: DE (74 Cor. 3). But CD-DE, .. Ad=de. In the same manner it may be shewn, that de=ef=fg=gB.

(3) Another still more ingenious method, which avoids entirely the drawing of any parallel line, is as follows:

From one extremity A of the given line draw an indefinite straight line, making any acute angle with AB; along which make six equal steps with the compasses, marking the 4th,

5th, and 6th, by the letters C, D, E. Join EB, and produce it to F, making BF=EB. Then join FC, intersecting AB in G, and BG shall be the 5th part of AB; so that taking this distance, BG, in the compasses and step

A

ping along AB from either A or B, the given line is divided as required.

For, joining BD, since ED = DC, and EB=BF, .. the sides EC, EF of the triangle ECF are divided proportionally in the points D and B; and .. BD is parallel to CF. Again, in the triangle ABD, since GC is parallel to BD, AG: AB :: AC: AD (70 Cor.); but