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For the greater segment is to the less, as the less to their difference; the less to the difference, as the difference to the difference between it and the less, and so on continually; no remainder being ever found, which is contained exactly in the preceding one. See page 270.

It thus appears, that the ratio of the parts of a line divided in this manner, cannot be exactly expressed by any two numbers. It may be approximated, however, as nearly as we please, by means of the following series of numbers, the successive terms of which are each equal to the sum of the two immediately preceding it, as is the case in a reversed order in the series of lines obtained in the scholium to the eleventh of the second book: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, &c.

Thus the parts are nearly as 3 to 5, more nearly as 5 to 8, more nearly still as 8 to 13, and so on.

PROP. D.

IF R be put to denote the radius of the circumscribed circle, p the perpendicular, S the area, and a, b, c the sides opposite to the angles A, B, C, we have from this proposition bc 2Rp. Multiply by a: then abc 2Rpa, or abc= 4RS, because pa 28. Hence, if the three sides of a triangle be given in numbers, we may compute the radius of the circumscribed circle, by dividing the continued product of the three sides by four times the area, computed (APP. IV. 29.) from the sides.

Hence also it appears (APP. V. 4.) that a parallelepiped, having for its length, breadth, and altitude, the three sides of a triangle, is equal to a prism having the triangle for its base, and twice the diameter of the circumscribed triangle for its altitude.

ELEMENTS, BOOK XI.

DEF. VIII.

INSTEAD of this definition the following is given in Dr. Simson's edition: "Similar solid figures are such as have all their solid angles equal, each to each, and which are contained by the same number of similar planes." This, instead of being a definition of similar solid figures in general, is only a definition of those which are contained by planes, and it should be restricted accordingly.

PROP. IV.

A SIMPLE and easy demonstration of this proposition has been substituted for Euclid's. It is the same in substance as that which has been given by Professor Wallace of Edinburgh, in the late editions of Playfair's Geometry.

PROP. VI.

THIS may also be demonstrated in the following manner. Draw DE of any length perpendicular to BD, and join AE, BE, AD. Then AE AB2 + BE2 = AB2 + BD2 + DE2 = AD2 + DE; wherefore (I. 48.) ADE is a right angle. The rest of the proof is the same as in the text.

PROP. XXIII.

By means of a lemma, a new demonstration of this proposition is given, which is much shorter and easier than that of Euclid.

PROP. XXV.

In this proposition and many others, Dr. Simson, rendering the Greek literally, uses the expression "solid parallelepiped." In this edition, the word "solid," is omitted as superfluous, in this and similar instances.

In connexion with this, it may be mentioned, that in the 16th proposition of the first book, and many others, I have used the phrase, remote angle, instead of opposite angle, confining the latter expression to the case in which an angle is mentioned in connexion with a straight line which subtends it. I have also, after Dr. Simson, avoided the use of the expression right line; adhering in both this and the preceding instance, to the principle, that in the language of science, as well as in other cases, words should be used as nearly as possible, in their ordinary and obvious meaning.

PROP. B.

WHEN, according to the hypothesis of this proposition, the solid angles at A and B are contained by plane angles which are not only equal, each to each, but which are similarly situated, so that if the planes CAD, FBG be made to coincide, the edges AE, BH lie on the same side of the coincident planes, the solid angles wholly coincide, and are therefore equal. But if the plane angles forming the two solid angles be arranged in a reversed order, so that when CAD, FBG are made to coincide, the edges AE, BH will be on opposite sides of the coincident planes, the solid angles will not coincide. Still, however, by proposition A., the inclinations of the planes forming the solid angles will be equal, and we

can have no hesitation in admitting, that the solid angles are themselves equal. Such solid angles as these, that is, such as are equal, and yet from lying in opposite directions, cannot coincide, have been denominated by Legendre, angles equal by symmetry, or simply symmetrical angles.

On the same principle there may be symmetrical solids; such, for instance, as two pyramids similarly described on opposite sides of the same base, and which though obviously equal, cannot be made to coincide by superposition.

Between plane figures and solids, there is a striking difference with regard to the present subject of consideration, as will appear by a simple example. If a perpendicular be drawn from the vertex of an isosceles plane triangle to the base, one of the parts into which the proposed triangle is thus divided, may be inverted and made to coincide exactly with the other. If, however, through the point in which the perpendicular cuts the base, a straight line be drawn perpendicular to the plane of the triangle, and through any point of that perpendicular straight lines be drawn to the several angles, a pyramid will be formed which will evidently be bisected by the plane passing through the perpendicular of the triangle and the perpendicular to its plane; and yet the halves of the pyramid cannot be made to coincide.

The principle on which we infer the equality of such solids, is, as has been remarked by Playfair, that the conditions which determine the magnitude of one of them, are the same as those which determine the magnitude of the other.

It is easy to see from these considerations, that, as was first shown by Dr. Simson, solids may be unequal, though contained by the same number of equal and similar planes. Thus, as a simple example, suppose two symmetrical pyramids described on opposite

The case is different with regard to an isosceles spherical triangle; as, while the perpendicular will evidently divide it, as well as the plane triangle into two triangles, having equal sides, equal angles, and equal areas, those triangles cannot be made to coincide, unless the sides of the isosceles triangle be quadrants.

It may be worth remarking, that, if without inverting either of the parts into which the perpendicular divides an isosceles plane triangle, the vertical angles be made to coincide, the parts of the triangles which do not coincide, will be two triangles having their sides and angles equal, each to each that if the vertical angles of these be applied to one another in a similar manner, the parts which do not coincide will also be triangles, having equal sides and angles, and so on; and if this process be continued, two triangles will at length remain, which will exactly coincide, if the angles of the proposed isosceles triangle be commensurable with a right angle: while if they be not, the process may be continued, till the triangles, which do not coincide, will be less than any assigned space. The proof of this is easy and it will be readily seen that there may be an analogous subdivision in case of an isosceles spherical triangle.

It may also be remarked, that if about any two such plane triangles, circles be described, and straight lines be drawn from their centres to the several angles, these lines and the sides of the triangles will form triangles, which will exactly coincide with each other: and the same will hold regarding spherical triangles, if arcs of great circles be drawn from the several angles to the nearer pole of the circumscribed circle.

sides of the same base: then, if on the same base another pyramid be described, lying wholly without the one of these, which is on the same side of the base with it, the whole solid thus formed, and the solid which remains when both the symmetrical pyramids are taken away, are evidently contained by the same number of equal and similar planes; while at the same time one of them is only a part of the other. It is plain also, that while the solid angles of these two unequal solids, are respectively contained by the same number of equal plane angles, some of these solid angles are unequal. No solid angles, indeed, except trihedral ones, are determined without more data than the plane angles which contain them.

The nature and relations of solid angles are well illustrated by means of spherical trigonometry; since, if a sphere be described from the vertex of a solid angle as centre, the solid angle will be measured by the portion of the spherical surface bounded by the planes forming the angle. On this subject the reader may consult Gregory's Trigonometry, Chap. VI. Section 6.

PROP. XXXV.

THE demonstration of this proposition is greatly shortened and simplified, chiefly by means of a corollary which has been annexed to proposition B.

APPENDIX, BOOK I.

PROP. VI.

THIS proposition has been inserted on account of its utility in the solution of many problems regarding triangles.

PROP. XX.

Of this proposition, which is the 10th of the thirteenth book of Euclid, I have given a demonstration which I believe to be new, and which is much easier than that of Euclid.

PROP. XXVI.

THIS curious proposition I have not met with, except in the "Dublin Problems," published in 1823, where it is inserted without demonstration. The following is an outline of a very easy and neat proof it by Mr. Adam D. Glasgow of Belfast, a former student of mine of great taste and talent for mathematical pursuits :

If circles be described about the triangles ABD, ACE, and if straight lines be drawn from A, B, C to their intersection, which suppose to be marked with L, the angles ALB, ALC will be

each two-thirds of two right angles, being (III. 22.) supplements of D and E and hence, since there remain two-thirds of two right angles for the angle BLC, it follows (APP. I. 4.) that the circle described about BCF will also pass through L. It is also plain, (see the second note to the tenth proposition of the first book,) that GK cuts AL perpendicularly, suppose in M; and that GH cuts BL perpendicularly, suppose in N. Then, since the angles at M and N are right angles, a circle (APP. I. 4.) may be described about GMLN: and therefore since MLN is two-thirds of two right angles, MGN (III. 22.) is one-third of two right angles. In the same manner it may be proved that GHK, HKG are each a third of two right angles: and hence (I. 6. cor.) GHK is an equilateral triangle.

PROP. XXXV. AND XXXVI.

THESE two propositions might be otherwise enunciated thus: "Given the sum of two straight lines, and the mean proportional between them; to determine them :" and "Given the difference of two straight lines, and the mean proportional between them; to determine them."

PROP. XLI. AND XLII.

In the year 1774, the interesting problem mentioned in the corollary to proposition XLI. was published, without solution, by the Rev. Mr. Lawson. The method of solving it was soon after discovered by Dr. Charles Hutton, and was acknowledged by Mr. Lawson to be on the same principle as his own. See Hutton's Tracts, Vol. I., page 254, and Vol. III., page 378; where also information will be found regarding proposition XLII.; the best mode of solving which seems also to be due to Dr. Hutton.

PROP. XLIII.

THIS proposition which is curious, as affording the first instance of the quadrature of a curvilineal space, is generally supposed to have been discovered by Hippocrates of Chios, an ingenious mathematician, who flourished about 450 years before Christ. For additional information respecting lunes, see Hutton's translation of Ozanam and Montucla's Recreations, Vol. I.

APPENDIX, BOOK II.

THE second book of the Appendix contains, in a small compass, much interesting matter. It commences with solutions of the "tangencies;" and it then introduces the student successively to the consideration of loci, of porisms, and of isoperimetrical figures.

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