Let ACD be the given circle:

it is required to inscribe a regular decagon in ACD.

Find the centre O, and draw

any

radius OA.

Divide OA at B, so that the rectangle contained by OA and AB may be equal to the square on OB,

II. Prob. 6. with centre A and radius equal to OB draw a circle cutting the circumference of ACD at C, join AC:

then shall AC be the side of a regular decagon inscribed in ACD.

Join OC, BC, and draw a circle through the three points O, B, C. III. Prob. 6.

Then, because the square on AC is equal to the rectangle contained by AO and AB,

therefore AC touches the circle OBC;

Constr. III. 28, Cor. 4.

therefore the angle ACB is equal to the angle BOC in the alternate segment of the circle;

III. 23.

therefore the whole angle ACO is equal to the sum of the angles BOC and BCO,

and therefore to the exterior angle ABC.

Ax. d.

I. 25.

But, because OA is equal to OC;

therefore the angle OAC is equal to the angle OCA,

and therefore to the angle ABC,

therefore BC is equal to AC,

and therefore to OB;

2.7.

Ax. c.

1. 8.

Ax. c.

I. 7.

therefore the angle BCO is equal to the angle BOC.

Hence the angle ACO is equal to the sum of two angles each of which has been proved equal to the angle AOC,

therefore also the angle OAC is equal to twice the angle AOC, and therefore AOC is a fifth-part of the sum of the angles of the triangle OAC,

and therefore a tenth part of four right angles;

and therefore the minor arc AC is a tenth part of the whole circumference ACD. III. 4.

Divide the whole circumference ACD into arcs equal to the minor arc AC, and draw the chords of these arcs; the polygon so formed is the regular inscribed decagon required. III. 26.

Also, by drawing tangents at the points at which the circumference is divided, a regular circumscribed decagon is obtained. III. 26.

By taking alternate points of division the circumference is divided into five equal parts, and the regular inscribed and circumscribed pentagons may be drawn.

repeatedly bisecting the arcs equal to AC, and drawing chords.

and tangents at the points of division.

Q.E.F.

Ex. 155. If the two circles intersect again at D, the chord CD is equal to AC. Also OB, BC, CD are three sides of a regular pentagon inscribed in the circle OBC.

Ex. 156. Describe an isosceles triangle having each of the angles at the base one-third of the vertical angle.

*Ex. 157. Divide a right angle into five equal parts.

Ex. 158. Shew that each diagonal of a regular pentagon is parallel to one of the sides.

PROB. 14. To inscribe in a circle a regular quindecagon, and thence to circumscribe a regular quindecagon about a circle; also to inscribe in, or to circumscribe about, a given circle regular figures of 30, 60, 120... sides.

Let ADB be the given circle:

it is required to inscribe a regular quindecagon in ADB.

Find the centre O, and draw any radius OA. Divide OA at C so that the square on AC may be equal to the

rectangle contained by AO and OC;

II. Prob. 6.

with centre A and radii AO and AC describe circles cutting the circumference of the circle ADB at B and D ;

join BD:

then shall BD be the side of a regular quindecagon inscribed in ADB.

For AB is the side of a regular hexagon inscribed in ADB,

III. Prob. 12.

III. Prob. 13.

and AD the side of a regular inscribed decagon; therefore of such parts as the whole circumference contains thirty, the arc AB contains five, and the arc AD three,

and therefore the arc BD contains two;

therefore the arc BD is a fifteenth part of the whole circumference.

Divide the whole circumference into arcs equal to the arc BD, and draw the chords of these arcs, the polygon so formed is the regular inscribed quindecagon required. III. 26.

Also by drawing tangents at the points at which the circumference is divided, a regular circumscribed quindecagon is obtained. III. 26.

Figures of 30, 60, 120 ... sides may be obtained by repeatedly bisecting the arcs equal to BD, and drawing chords, and tangents at the points of division.

Q.E.F.

EXERCISES.

159. Divide a given straight line so that the rectangle contained by its segments shall be equal to a given square not greater than that on half the line.

160.

Find a point in the base of a triangle such that the square on the straight line drawn to the vertex may be equal to the rectangle contained by the segments of the line. Distinguish the cases in which there are two solutions, one, or none.

161. If one side of a pentagon be produced, trisect the external angle.

162. If with one of the angular points of a regular pentagon as centre and one of its diagonals as radius a circle be described; a side of the pentagon will be equal to the side of the regular decagon inscribed in the circle.

163. The centre of each of two equal circles lies on the circumference of the other. Shew that the square on the common chord is three times that on the radius.

164. If from a point without a circle there be drawn two straight lines, one of which is perpendicular to a diameter, and the other cuts the circle: the square on the perpendicular is equal to the sum or difference of the rectangle contained by the whole cutting line and the part without the circle and the rectangle contained by the segments of the diameter according as the diameter is divided internally or externally by the perpendicular.