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

Corol. The inscribed circle touches the middle of the sides of the polygon.

PROBLEM XXXII.

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 or will be the radius.

A

F

B

E

D

For the perpendiculars to the tangents AF, AG, pass through the centre (cor. th. 47); and the inscribed circle touches the middle points 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, oc, will also be equal (cor. th. 45). There. fore 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 XXXIII.

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 oD, will be the radius.

For, draw оB, OA, OE, &c. to the angular points of the given polygon. Then

BA

the triangle OCD is isosceles, having the angles at c and D equal, being the halves of the equal angles of the polygon BCD, CDE; therefore 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 ocp, OCB, also equal, will be identical (th. 1), and have their third sides no, OD, equal. In like manner it is shown, that all the lines OA, 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 circum. scribe the polygon.

PROBLEMS.

PROBLEM XXXIV.

To make a square equal to the sum of two or more given

squares.

LET AB and AC be the sides of two

Draw two indefinite given squares. lines AP, AQ, at right angles to each other; in which place the sides ab, ac, of the given squares; join BC; then a square described on BC will be equal to the sum of the two squares described on AB and AC (th. 34).

In the same manner, a square may be made equal to the sum of three or more given squares. For, if ab, ac, ad, be taken as the sides of the given squares, then, making aɛ=BC, AD = AD, and drawing ne, it is evident that the square on DE will be equal to the sum of the three squares on ab, ac, ad. And so on for more squares.

PROBLEM XXXV.

To make a square equal to the difference of two given

squares.

LET AB and AC, taken in the same straight line, be equal to the sides of the two given squares.--From the centre A, with the distance AB, describe a circle; and make CD perpendicular to AB, meet

A CB

ing the circumference in D: so shall a square described on CD be equal to AD-AC2, or AB-AC, as required (cor. th. 34).

PROBLEM XXXVI.

To make a triangle equal to a given quadrangle ABCD.

DRAW the diagonal ac, and parallel to it DE, meeting BA produced at E, and join CE; then will the triangle CEB be equal to the given quadrilateral ABCD.

For, the two triangles ACE, ACD, being on the same base AC, and between

D C

E A

B

the same parallels AC, DE, are equal (th. 25); therefore, if ABC be added to each, it will make вCE equal to ABCD (ax. 2).

VOL. I.

48

PROBLEM XXXVII.

To make a triangle equal to a given pentagon ABCDE.

DRAW DA and DB, and also EF, CG, parallel to them, meeting AB produced at and G; then draw DF and DG; so shall the triangle DFG be equal to the given pentagon abcde.

For the triangle DFA = DEA, and the triangle DGB = DCB (th. 25); therefore, by adding DAB to the equals,

E

D

FA

B G

the sums are equal (ax. 2), that is, DAB+ DAF + DBG= DAB + DAE + DBC, or the triangle DFG = to the pentagon ABCDE.

PROBLEM XXXVIII.

To make a rectangle equal to a given triangle ABC.

BISECT the base AB in D: then raise DE and Bг perpendicular to AB, and meeting CF parallel to AB, at E and F : so shall or be the rectangle equal to the given triangle ABC (by cor. 2, th. 26).

A

CE F

PROBLEM XXXIX.

To make a square equal to a given rectangle ABCD.

PRODUCE one side AB, till BE be

equal to the other side BC.

On AE as a diameter describe a circle, meeting BC produced at F: then will BF be the side of the square BFGн, equal to the given rectangle BD, as required; as appears by cor. th. 87, and th. 77.

G

F

D

A H

BE

APPLICATION OF ALGEBRA

ΤΟ

GEOMETRY.

WHEN it is proposed to resolve a geometrical problem algebraically, or by algebra, it is proper, in the first place, to draw a figure that shall represent the several parts or conditions of the problem, and to suppose that figure to be the true one. Then having considered attentively the nature of the problem, the figure is next to be prepared for a solu tion, if necessary, by producing or drawing such lines in it as appear most conducive to that end. This done, the usual symbols or letters, for known and unknown quantities, are employed to denote the several parts of the figure, both the known and unknown parts, or as many of them as necessary, as also such unknown line or lines as may be easiest found, whether required or not. Then proceed to the operation, by observing the relations that the several parts of the figure have to each other; from which, and the proper theorems in the foregoing elements of geometry, make out as many equations independent of each other, as there are unknown quantities employed in them: the resolution of which equations, in the same manner as in arithmetical problems, will determine the unknown quantities, and resolve the problem proposed.

As no general rule can be given for drawing the lines, and selecting the fittest quantities to substitute for, so as always to bring out the most simple conclusions, because different problems require different modes of solution; the best way to gain experience, is to try the solution of the same problem in different ways, and then apply that which succeeds best, to other cases of the same kind, when they afterwards occur. The following particular directions, however, may be of

some use.

1st, In preparing the figure, by drawing lines, let them be either parallel or perpendicular to other lines in the figure, or so as to form similar triangles. And if an angle be given, it will be proper to let the perpendicular be opposite to that angle, and to fall from one end of a given line, if possible.

2d, In selecting the quantities proper to substitute for, those are to be chosen, whether required or not, which lie nearest the known or given parts of the figure, and by means of which the next adjacent parts may be expressed by addition and subtraction only, without using surds.

3d, When two lines or quantities are alike related to other parts of the figure or problem, the best way is, not to make use of either of them separately, but to substitute for their sum, or difference, or rectangle, or the sum of their alternate quotients, or for some line or lines, in the figure, to which they have both the same relation.

4th, When the area, or the perimeter, of a figure is given, or such parts of it as have only a remote relation to the parts required it is sometimes of use to assume another figure similar to the proposed one, having one side equal to unity, or some other known quantity. For, hence the other parts of the figure may be found, by the known proportions of the like sides, or parts, and so an equation be obtained. For examples, take the following problems.

PROBLEM I.

In a right-angled triangle, having given the base (3), and the sum of the hypothenuse and perpendicular (9); to find both these two sides.

LET ABC represent the proposed triangle right-angled at B. Put the base AB = 3 = b, and the sum AC+BC of the hypothenuse and perpendicular 9 = s; also, let x denote the hypothenuse ac, and y the perpendicular BC.

=

A

B

y,

s2-2sy + y2 = y2 + b3,

gives Taking away y'on both sides leaves s2-2sy = b2,

By transpos. 2sy and b2, gives

And dividing by 2s, gives

Hence x=8- y= 5 = AC.

83-bi = 2sy,

= y = 4 = BC.

2$

N. B. In this solution, and the following ones, the nota. tion is made by using as many unknown letters, z and y, as