each polygon will have a common vertex, and their sum will be equal to the polygon; or rather we can suppose that all the triangles formed in a polygon have for a common base a side of the polygon, and for vertices those of the different angles opposite to this base. In each case the number of triangles formed being n-2, the conditions of their similitude will be equal to the number 2n-4; and the definition will contain nothing superfluous. This new definition being adopted, the ancient one will become a theorem, which may be demonstrated immediately.

If the definition of similar rectilineal figures is imperfect in books of elements, that of similar solid polyedrons is still more so. In Euclid this definition depends upon a theorem not demonstrated; in other authors it has the inconvenience of being very redundant; we have, therefore, rejected these definitions of similar solids.*

The definition of a perpendicular to a plane may be regarded as a theorem; that of the inclination of two planes also requires to be supported by reasoning; the same may be said of several others. It is on this account that, while we have placed the definitions according to ancient usage, we have taken care to refer to propositions where they are demonstrated; sometimes we have merely added a brief explanation, which appeared sufficient.

The angle formed by the meeting of two planes, and the solid angle formed by the meeting of several planes in the same point, are distinct kinds of magnitudes, to which it would be well, perhaps, to give particular names. Without this it is difficult to avoid obscurity and circumlocutions in speaking of the arrangement of planes which compose the surface of a polyedron; and, as the theory of solids has been little cultivated hitherto, there is less inconvenience in introducing new expressions, where they are requi.ed by the nature of the subject.

I should propose to give the name of wedge to the angle formed by two planes; the edge or height of the wedge would be the common intersection of the two planes. The wedge would be designated by four letters, of which the two middle ones would answer to the edge. A right wedge, then, would be the angle formed by two planes perpendicular to each other. Four right wedges would fill all the solid angular space about a given line. This new denomination would not prevent the wedge always having for its measure the angle formed by two lines drawn from the same point, the one in one of

*The author here refers to a distinct note on the equality and similitude of polyedrons, not given in this translation.

the planes and the other in the other, perpendicularly to the edge or common intersection.

II.

By the Translator.

THE improvements referred to in the preceding note, so far as they have been adopted by the author, have been carefully preserved in the translation. Indeed, it has been found necessary, in a few instances, to use English words in a sense somewhat different from their ordinary acceptation. The word polygon is generally restricted to figures of more than four sides. It is used in this work with the latitude of the original word polygone to stand for rectilineal figures generally; and polyedron is adopted in a similar manner for solids. Quadrilateral is employed as a general name for four-sided figures. The word losenge is rendered by rhombus, and trapezé by trapezoid, the English words, as they are commonly used, corresponding to the French. The perpendicular let fall from the centre of a regular polygon upon one of its sides is called in the original apothéme. It occurs but a few times, and as there is no English word answering to it, it is rendered by a periphrasis, or simply by the word perpendicular. The portion of the surface of a sphere comprehended between the semicircumferences of two great circles is denoted in the original by fuseau; Dr. Hutton uses the word lune in the same sense; others have employed lunary surface; as lune properly stands for the surface comprehended between two unequal circular curves, the latter denomination was thought the least exceptionable, and is adopted in the translation.

QUESTIONS IN GEOMETRY,

INTENDED AS AN EXERCISE FOR THE LEARNER

Q. I. From two given points A and B, on the same side of a line DE, Fig, 293. given in position, to draw two lines AP, PB, which shall meet in DE, and make equal angles with it. (34. 36.)

Q. II. From two given points A and B, to draw two equal straight lines Fig. 294. AE, BE, which shall meet in the same point of a line CD. (36 or 55.)

Q. III. Through a given point P, to draw a line FE, which shall make Fig. 295. equal angles with two given straight lines BE, CF. (38.)

Q. IV. If, from two points A and B, on the same side of a given line DE, Fig. 296 two straight lines AP, PB, be drawn, making equal angles with DE, AP and PB, will be together less than the sum of any other two lines AG, GB, drawn from A and B to any other point G in the line DE. (38. 40.)

Q. V. If the three sides of a triangle be bisected, the perpendiculars Fig. 297. drawn to the sides at the three points of bisection, will meet in the same point.

Let the sides of the triangle ABC be bisected in the points D, E, F. Draw the perpendiculars EG, FG, meeting in G. The perpendicular at D also passes through G. (36. 48.)

Q. VI. To divide a right angle into three equal parts.-Let ACB be Fig. 298. a right angle. Take CA of any magnitude, and erect upon it an equilateral triangle. (62.)

Q. VII. Let ABC be an equilateral triangle. If the angles CBA and Fig. 299. CAB be bisected by the lines AD and BD, meeting in a point D, and DE, DF, be drawn parallel to the sides CA and CB respectively, the line AB will be divided into three equal parts at the points E and F. (76. 48.)

Q. VIII. Any side of a triangle is greater than the difference of the Fig. 300 other two.

Let AC be greater than AB, and cut off AD equal to AB.

(45.63.)

Q. IX. If, from B, the vertex of the triangle ABC, BE be drawn perpen- Fig. 301 dicular to the base, and BD bisecting the angle ABC, the difference of the angles BAC and BCA is double the angle EBD. (57.)

Q. X. If, from B one of the equal angles of an isosceles triangle, any Fig. 302 line BD be drawn to the opposite side, and from the same point B a line BE

be drawn to the opposite side produced, so that DE shall be equal to DB, the angle ABD will be double the angle CBE. (63. 45.)

Geom. 29

Fig. 303

Fig, 304.

Fig. 305.

Fig. 306.

Fig. 307.

Fig 308.

Fig. 309.

Fig. 310.

Fig. 311.

Fig. 3.2.

Q. XI. If, from the extremity C of the base BC of an isosceles triangle, a line CD equal to AC be drawn to meet the opposite side AB, produced if necessary, the angle DCE, formed by this line and the base produced, will be equal to three times the angle ABC. (45.63.)

Q XII. The sum of the sides of an isosceles triangle is less than the sum of the sides of any other triangle on the same base and between the same parallels.

Let ACB be an isosceles triangle, and ADB any other triangle on the same base AB and between the same parallels AB and ED, AC and CB to gether will be less than AD and DB. (76. Q. IV.)

Q. XIII. If the three angles of the triangle ABC be bisected by the lines AD, BD, CD, and BD be produced to E, and if from D, DF be drawn perpendicular to AC, the angle ADE will be equal to CDF. (57.)

Q. XIV. To bisect a parallelogram by a line drawn from a point in one of its sides.

Let ABCD be a parallelogram, and P a given point in the side AB. Draw the diagonal BD, which bisects the parallelogram. Bisect BD in F, and through P and F draw PFE. PE bisects the parallelogram.

Q. XV. If, in the sides of a square ABCD, four points E, F, G, H, be taken, one in each side, at equal distances from the four angles, the figure EFGH, contained by the straight lines which join these points, will be a square.

Q. XVI. If the sides of the square described upon the hypothenuse of a right angled triangle be produced to meet the sides (produced if necessary) of the squares described upon the sides of the triangle, they will cut off triangles equiangular, and equal to the given triangle.

Let DB, EC, the sides of the square described on BC, the hypothenuse of a right angled triangle ABC, be produced to meet the sides of the squares described upon BA, AC, in K and L; the triangles BFK, CIL, cut off by them, are equiangular, and equal to ABC.

Q. XVII. To inscribe a square in a given right angled isosceles triangle.

Divide the base AC of the right angled isosceles triangle into three equal parts.

Q. XVIII. If the sides of an equilateral and equiangular pentagon be produced to meet, the angles formed by these lines are together equal to two right angles.

Let the sides of the equilateral and equiangular pentagon ABCDE be produced to meet in the points F, G, H, I, K; the angles at these points are together equal to two right angles. (63.)

Q. XIX. If the sides of an equilateral and equiangular hexagon, ABCDEF, be produced to meet in the points G, H, I, K, L, M, the angles at these points are together equal to four right angles.

Q. XX. The perimeter of an isosceles triangle is greater than the perimeter of a rectangular parallelogram which is of the same altitude, and equal to the given triangle.

Let ABC be an isosceles triangle whose base is BC. Draw AE perpendicular to BC, and therefore bisecting it, and draw AD, CD, parallel respectively to BC, AE; then DE is a rectangular parallelogram of the same altitude with, and equal to, the triangle ABC. The perimeter of ABC is greater than that of DE.