and from B and D, with the same radius, describe arcs cutting in F. Join AE, ED, DF, FB.

These, with AB, form a regular pentagon described on AB.

Problem 62.

215. To make a regular hexagon on a given straight line.

Let AB, fig. 57, be the given straight line. From A and B, with radius AB, describe arcs intersecting in C. With C as centre, and the same radius, describe a circle. It will pass through A and B. With the same radius, stepping round from A, mark the points G, F, E, D in the circumference. These, with A and B, will be angular points in the required polygon, which will be formed by straight lines joining the adjacent points.

216. Note 1.-The side of any hexagon is equal to the radius of the circumscribed circle. Hence

To inscribe a hexagon in a circle.

With the radius of the circle, step round the circumference, marking six successive points. The lines joining these points will be a hexagon inscribed in the circle.

217. Note 2.-To inscribe an equilateral triangle in a circle. Mark, as just described, the points in the circumference for a hexagon. Lines joining alternate points will form an equilateral triangle inscribed in a circle; as, AF, AD, FD, fig. 57.

218. Note 3.-To inscribe a dodecagon (a regular twelve-sided figure) in a circle. Mark, as in Note 1, two adjacent points in the circumference for a hexagon. Bisect the arc thus found, as AG in H, fig. 57.

The distance AH, applied to mark points round the circumference, will give the angular points of a dodecagon inscribed in the circle.

E

Problem 63.

219. To make an octagon on a given straight line. Let AB be the given straight line.

Draw AC and

BD perpendicular to AB, and produce AB both ways, to E and F. Bisect the angles CAE and DBF by the lines AG and BH, making each equal to AB. Through G and H, draw GK and HL parallel to AC or BD, and each equal to AB. From the centres K and L, with the radius AB, cut AC in C and BD in D. Join KC, CD, DL. ABHLDCKG is the required octagon.

EXERCISES ON SEVERAL OF THE PROBLEMS.

These exercises are designed to train to careful work, and at the same time to interest and teach a few useful geometrical truths.

220. Problem 2.-Draw any circle. With the legs of the compasses at the same distance used in drawing it (i.e., with the same radius), set one foot on any point

in the circumference, and with the other foot mark another point in the circumference. Keeping this last foot fixed, bring round the foot first set down, and mark a third point in the circumference; and so on, stepping round with each foot alternately on a point in the circumference. The sixth step should bring the foot of the compasses back to the point first taken.

The circumference will thus be divided into six arcs of 60° each.

Join each point by straight lines to the points next it on each side. Each line thus drawn will be the chord of 60°.

It is thus plain that, in every circle, the chord of 60° is equal to the radius.

The rectilineal figure formed by these six chords is a regular hexagon, or regular six-sided figure.

221. Draw a circle. From any point in the circumference, with the same radius, draw an arc within the circle, its ends being in the circumference. Step round the circle, drawing similar arcs from the ends of those previously described, till six such arcs have been drawn.

If correctly done, these arcs will all pass through the centre, and their twelve ends will meet, two and two, in six points in the circumference, dividing it into six arcs of 60° each. Carefully done, this forms a very elegant figure.

222. Problem 3.-Find three or more points, each equidistant from the ends of any straight line. Join the equidistant points. If correctly done, they will be in the same straight line; and that line, sufficiently produced, will bisect the given straight line, and be at right angles to it.

This illustrates a geometrical truth, already pointed out (70).

223. Problem 4.-Bisect each of the angles of any

:

triangle the bisecting lines should intersect each other in the same point.

224. Draw any two straight lines cutting each other. Bisect each of the four angles at the point of intersection. Each bisecting line should be at right angles to the adjacent bisecting lines, and continuous with the opposite one.

225. Bisect the angle between the equal sides of an isosceles triangle, or any angle of an equilateral triangle. The bisecting line will bisect the base, and be at right angles to it. See Note 2, Problem 4. If DE be drawn, BDE is an isosceles or equilateral triangle.

226. Problem 5.--If the operation has been accurately done, the bisecting line, CD, will be at right angles to the line to be bisected, AB.

227. Bisect the base of an isosceles triangle, or any side of an equilateral triangle. The straight line from the point of bisection to the opposite angle will bisect that angle, and be at right angles to the side bisected.

228. Bisect any two sides of a triangle. The line joining the points of bisection will be parallel to the third side, and equal to half of it.

229. Bisect the four sides of any quadrilateral. The line joining the points of bisection of the adjacent sides will form a parallelogram.

230. Problems 6, 7.-Bisect the base of an isosceles triangle, or any side of an equilateral triangle. Through the point of bisection, draw inwards a perpendicular to the bisected line. The perpendicular will pass through the opposite angular point and bisect the angle there.

231. Bisect any straight line. From the point of bisection, draw a perpendicular both ways. Every point of this perpendicular should be equidistant from the ends of the line.

232. Bisect any angle. From any point in the bisecting line, draw perpendiculars to the sides. They should be equal.

Note. The straight line bisecting an angle is the locus of points equidistant from its sides; that is, any point equidistant from the sides is in that line.

233. From the angular points of any triangle, draw perpendiculars to the opposite sides. They should meet in one point, within the triangle, or without it when produced.

234. Bisect the sides of any triangle. At the points of bisection, draw perpendiculars. They will meet in one point, which will be equidistant from the angular points of the triangle. If the triangle is acute angled, the perpendiculars will meet within it; if right angled, in the middle point of the hypotenuse; if obtuse angled, without the triangle.

235. Bisect the three angles of a triangle. From the point in which the bisecting lines meet, draw perpendiculars to the sides. These perpendiculars should be equal, and a circle from their point of intersection as centre, with one of them as radius, should pass through the ends of all the perpendiculars, and touch each side without cutting it.

236. Problem 8.—Let ABC be any angle less than a right angle. At C, on the same side of BC, make an angle equal to ABC, with its other side leaning towards BA. The sides of the two equal angles will meet, if produced sufficiently, and an isosceles triangle will be formed, the sides opposite the equal angles being equal.

Next, make the angle at C equal to ABC to lie the other way, producing BC. The side of the angle at C will now be parallel to BA; illustrating the geometrical truth mentioned in Par. 188.

237. Problem 9.-Trisect any right angle, ABE, by the lines BC, next to BA, and BD next to BE. At A and E, in the lines AB, EB, make angles, each equal to ABD or CBE; that is, two-thirds of a right angle. Let the trisecting lines be sufficiently produced to meet the