Next, if the base AB be given, by the second case of the foregoing propo sition, produce AB to C, so that AC X CB = AB2 then will AB be the base, and AC the length of each A of the two sides.

B

C

PROBLEM XXXVII

To describe a regular pentagon on a given line AB.

On AB erect the isosceles triangle ACB having each of the angles at the base double of its vertical angle, on AB again construct another isosceles triangle whose vertical angle AOB is double of ACB, and about the vertex O place the isosceles triangles AOD, DOC, COE, and EOB; these triangles, with AOB, will compose a regular pentagon.

A

E

For the angle AOB, being the double of ACB, which is the fifth part of two right angles, must be equal to the fifth part of four right angles; and consequently five angles, each of them equal to AOB, will adapt themselves about the point O. But the bases of those central triangles, and which form the sides of the pentagon, are all equal; and the angles at their bases being likewise equal, they are equal in the collective pairs which constitute the internal angles of the figure. It is therefore a regular pentagon.

PROBLEM XXXVIII.

To describe a hexagon upon a given line AB.

From A and B as centres, with AB as radius, describe arcs intersecting in O (fig. to the next problem). From O as a centre, with the same radius, describe a circle ABCDEF. Within this circle set off from B, the chords BC, CD, DE, EF, FA, in succession, each equal to AB: they will, together with AB, form the hexagon required.

The demonstration is analogous to that of the following problem.

PROBLEM XXXIX.

To inscribe a regular hexagon in a circle.

Apply the radius AO of the given circle as a chord, AB, BC, CD, &c. quite round the circumference, and it will complete the regular hexagon ABCDEF.

For, draw the radii AO, BO, CO, DO, EO, FO, completing six equal triangles; of which any one, as ABO, being equilateral (by constr.), its three angles are all equal (cor. 2, th. 3), and any one of them, as AOB, is one-third of the whole, or of two right angles (th. 17), or one-sixth of four right angles. of four right angles (cor. 4, th, 6).

E

But the whole circumference is the measure
Therefore the arc AB is one-sixth of the

circumference of the circle, and consequently its chord AB one side of an equilateral hexagon incribed in the circle. And the same of the other chords.

Cor. The side of a regular hexagon is equal to the radius of the circumscribing circle, or to the chord of one-sixth part of the circumference.

PROBLEM XL

On a given line AB to construct a regular octagon.

Bisect AB by the perpendicular CD, which make = CA or CB, join DA and DB, produce CD making DO = DA or DB, draw AO and BO, thus forming an angle equal to the half of ADB, and about the vertex O repeat the equal triangles AOB, AOE, EOF, FOG, GOH, HOI, IOK, and KOB to compose the octagon.

For the distances AD, BD are evidently equal; and because CA, CD, and CB are all equal, the angle ADB is contained in a semicircle, and is, therefore, a right angle. Consequently AOB is equal to the half of a right angle, and eight such angles will adapt themselves about the point O. Whence the figure BAEFGHIK, having eight equal sides and equal angles, is a regular octagon.

PROBLEM ILL

To divide the circumference of a given circle successively into 4, 8, 12, and 24 equal parts.

1. Insert the radius AB three times from A to D, E, and C; from the extremities of the diameter AC, and with a distance equal to the double chord AE, describe arcs intersecting in the point F; and from A, with the distance BF, cut the circumference on opposite sides at G and H: AG, GC, CH, and HA are quadrants.

2. From the point F with the radius AB, cut the circle in I and K, and from A and C inflect the chord AI from L and M; the circumference is divided into eight equal portions by the points A, I, G, K, C, M, H, and L.

3. The arc DG, on being repeated, will form twelve equal sections of the circumfe

rence,

4. The arc ID is the twenty-fourth part of the circumference.

Ky

E

B

H

PROBLEM XLII.

To divide the circumference of a given circle successively into 5, 10, and 20 equal parts.

Mark out the semicircumference ADEC by the triple insertion of the radius, from A and C with the double chord AE describe arcs intersecting in F, from

A with the distance BF cut the circle in G and H, inflect the chords GH and GI equal to the radius AB, and from the points H and I, with distance BF or AG, describe arcs intersecting in L.

For BL is the greater segment of the radius BH divided by a medial section; wherefore AL is equal to the side of the inscribed pentagon, and BL, to that of the decagon inscribed in the given circle. Hence AL may be inflected five times in the circumference, and BL ten times;

and consequently the arc MK, or the excess of the fourth above the fifth, is equal to the twentieth part of the whole circumference.

PROBLEM XLIII.

To describe a regular pentagon, hexagon, or octagon, aboïst a circle.

In the given circle inscribe a regular polygon of the same name or number of sides, as ABCDE, by one of the foregoing problems. Then to all its K angular points draw tangents (by prob. 22), and these will form the circumscribing polygon required.

For all the chords, or sides of the inscribed figure, AB, BC, &c., being equal; and all the radii OA, OB, &c., being equal; all the vertical angles about the point O are equal. But the angles OEF, OAF, OAG, OBG, made by the tangents and radii, are right angles; therefore OEF + OAF = two right angles, and OAG + OBG two right angles; consequently, also, AOE + AFE and AOB AGB = two right angles (cor. 2, th. 18). angles AOE+ AFE being = AOB + AGB, of which consequently, the remaining angles F and G are also equal. In the same manner it is shown, that all the angles F, G, H, I, K, are equal.

two right angles, Hence, then, the AOB is = AOE;

Again, the tangents from the same point FE, FA, are equal, as also the tangents AG, GB (cor. 2, th. 61); and the angles F and G of the isosceles triangles AFE, AGB, are equal, as well as their opposite sides AE, AB; consequently, those two triangles are identical (th. 1), and have their other sides EF, FA, AG, GB, all equal, and FG equal to the double of any one of them. In like manner it is shown, that all the other sides GH, HI, IK, KF, are equal to FG, or double of the tangents GB, BH, &c.

Hence, then, the circumscribed figure is both equilateral and equiangular; which was to be shown.

Cor-The inscribed circle touches the middle of the sides of the polygon.

PROBLEM XLIV.

To inscribe a circle in a regular polygon.

Bisect any two sides of the polygon by the perpendiculars GO, FO, and their intersection O will be the centre of the inscribed circle, and OG or OF will be the radius.

B

G

E

For the perpendiculars to the tangents AF, AG, pass through the centre (cor., th. 47); and the inscribed circle touches the middle point F, G, by the last corollary. Also, the two sides AG, AO, of the right-angled triangle AOG, being equal to the two sides AF, AO, of the right-angled triangle AOF, the third sides OF, OG, will also be equal (cor., th. 45). Therefore, the circle described with the centre O and radius OG will pass through F, and will touch the sides in the points G and F. And the same for all the other sides of the figure.

PROBLEM XLV.

To describe a circle about a regular polygon.

Bisect any two of the angles C and D with the lines CO, DO; then their intersection O will be the centre of the circumscribing circle; and OC, or O1), will be the radius.

For, draw OB, OA, OE, &c., to the angular points of the given polygon. Then the triangle OCD is isosceles, naving the angles at C and D equal, being the halves

of the equal angles of the polygon BCD, CDE; there

P

E

fore, their opposite sides CO, DO, are equal (th. 4). But the two triangles OCD, OCB, having the two sides OC, CD, equal to the two OC, CB, and the included angles OCD, OCB, also equal, will be identical (th. 1), and have their third sides BO, OD, equal. In like manner it is shown, that all the lines ÓA, OB, OC, OD, OE, are equal. Consequently, a circle described with the centre O and radius OA, will pass through all the other angular points, B, C, D, &c., and will circumscribe the polygon.

PROBLEM XLVI.

On a given line to construct a rectilinear figure similar to a given rectilinear figure.

Let abcde be the given rectilinear figure, and AB the side of the proposed similar figure that is similarly posited with ab.

Place AB in the prolongation of ab, or parallel to it. Draw AC, AD, AE, &c., parallel to ac, ad, ae, respectively. Draw BC parallel to be, meeting AC in C; CD

E

B

parallel to cd, and meeting AD in D; DE parallel to de, and meeting AE in E; and so on, till the figure is completed. Then ABCDE will be similar to abcde, from the nature of parallel lines and similar figures (th. 89).

MISCELLANEOUS EXERCISES IN PLANE GEOMETRY.

(1.) From two given points, to draw two equal straight lines, which shall meet in the same point of a line given in position.

(2.) From two given points, on the same side, or opposite sides of a line given in position, to draw two lines, which shall meet in that line, and make equal angles with it.

(3.) To trisect a given finite straight line.

(4.) If from the extremities of the diameter of a semicircle, perpendiculars De let fall on any line cutting the semicircle, the parts intercepted between those perpendiculars and the circumference are equal.

(5.) If on each side of any point in a circle any number of equal arcs be taken, and the extremities of each pair joined, the sum of the chords so drawn will be equal to the last chord produced to meet a line drawn from the given point through the extremity of the first arc.

(6.) If one circle touch another externally or internally, any straight line drawn through the point of contact will cut off similar segments.

(7.) If two circles touch each other, and also touch a straight line, the part of the line between the points of contact is a mean proportional between the diameters of the circles.

(8.) From two given points in the circumference of a given circle, to draw two lines to a point in the circumference, which shall cut a line given in position, so that the part of it intercepted by them may be equal to a given line.

(9.) If from any point within an equilateral triangle perpendiculars be drawn to the sides, they are, together, equal to a perpendicular drawn from any of the angles to the opposite side.

(10.) If the three sides of a triangle be bisected, the perpendiculars drawn to the sides, at the three points of bisection, will meet in the same point.

(11.) If from the three angles of a triangle lines be drawn to the points of bisection of the opposite sides, these lines intersect each other in the same point.

(12.) The three straight lines which bisect the three angles of a triangle, meet in the same point.

(13.) If from the angles of a triangle perpendiculars be drawn to the opposite sides, they will intersect in the same point.

(14.) If any two chords be drawn in a circle, to intersect at right angles, the sum of the squares of the four segments is equal to the square of the diameter of the circle.

(15.) In a given triangle to inscribe the greatest square.

16.) In a given triangle to inscribe a rectangle, whose sides shall have a given ratio.

DD