tion, and (4) a proof. The parts of the enunciation of a problem are the datum or data, and the requirement; the solution consists of the construction and the proof. In a theorem, the hypothesis and thesis make up the enunciation; there may be a construction, there is necessarily always a proof. d. Again, when the hypothesis of one proposition becomes in another that which is to be proved; and that which is proved in the former becomes the hypothesis of the latter, each of such propositions is called the converse of the other. e. There is the same sort of difference between a postulate and an axiom, as between a problem and a theorem. A postulate is simply a problem so easy that there cannot be any set of directions laid down for doing it; and that the mode of doing it must be known to every one; it is a request or demand which will at once be granted. Postulates, in fact, grant permission to use the rule and compass, and these only, in the construction of geometrical figures. An axiom is a self-evident proposition, that is, one the truth of which the mind at once admits when the terms in which it is stated are understood. The strict rule in geometry is, that nothing is taken for granted as an axiom which is capable of proof, and it is upon this principle that Euclid always proceeds. On this ground we have adopted as the twelfth axiom one which is a slight modification of that of Euclid, and presents the same truth in a more self-evident form. We admit, however, that objection may be taken to it; but we consider it the best way of escape out of a great difficulty. f. Postulates 1 and 2 are understood as not authorising the use of marked divisions on the rule, or straight-edge, employed in constructions. Postulate 2, viewed in connection with def. 5, shews that if two straight lines coincide, their continuations will also coincide (see Prop. 4, I., sch.) The third postulate, as stated by Euclid, does not authorise the transference of distances by the compasses, but merely the description of a circle from one end of a given line, and so that the circumference shall pass through the other. But this restriction has been abandoned by most geometers, and the compasses used to carry distances. In this way many problems are greatly simplified, complicated constructions being avoided. Thus Euclid's third Prop. requires the description of five circles and a triangle, and the drawing of three lines in assigned positions, while every one solves it in practice by transferring the distance by the compasses. The restriction in regard to carrying distances being abandoned, the formal statement of Props. 2 and 3 might have been omitted, and the constructions assumed as two additional postulates--a method adopted by many geometers. PROPOSITIONS. PROPOSITION I.--PROBLEM. To describe an equilateral triangle on a given finite straight line. Let AB be the given finite straight line; it is required to describe an equilateral triangle upon it. From the centre A, at the distance AB, describe (post. 3) the circle BCD, and from the centre B, at the distance BA, describe (post. 3) the circle ACE; and from the point C', in which the circles cut one D B Because the point A is the centre of the circle BCD, (I. Def. 30.) and because the point B is the centre of the circle ACE, BC is equal to BA; (I. Def. 30.) (Ax. 1.) therefore CA and CB are each of them equal to AB; but things which are equal to the same thing are equal to one another; therefore CA is equal to CB; and the three lines BA, AC, CB, are all equal, so that the triangle is equilateral; and it is described upon the given finite straight line AB. Q. E. F. Schol. It is manifest that lines drawn from C', the other point of intersection of the two circles, to the extremities of the given line, will also form with it an equilateral triangle. B PROP. II.-PROBLEM, From a given point to draw a straight line equal to a given straight line. | B A D Let A be the given point, and B the given straight line; it is required to draw from A a straight line equal to B. From the centre A, with the radius B, describe (post. 3) an arc of a circle, and from the centre A draw a straight line AD to any point D in the arc. Then because AD is equal (def. 30) to B, AD is the line required. Schol. An isosceles triangle, whose sides shall have a given length, may be described on a given finite straight line, after the manner of Prop. I., provided that the given length be greater than half the line on which the triangle is to be described. In this way Prop. I. is made more general, PROP. III.-PROBLEM. From the greater of two given straight lines to cut off a part equal to the less. Let AB be the greater line and C the less; it is required to cut off from AB a part equal to C. A From the centre A, with B the radius C, describe (post. 3) an arc of a circle, cutting AB in D.-Then because AD is equal (def. 30) to C, AD is the part required.—The less line may also be produced to equality with the greater; and a line may be drawn equal to the sum of the two lines. These Propositions might have been omitted, as they are merely formal statements of the assumption which has been made, that distances may be transferred by the compasses. (See p. 16, f). PROP. IV. THEOREM. Two triangles are equal in every respect- 1. If two sides and the angle between them in the one triangle be respectively equal to two sides, and the angle between them in the other; 2. If two angles and the side between them in the one triangle be respectively equal to two angles, and the side between them in the other. Let ABC, DEF, be two triangles, in which 1. The side AB is equal to the side DE, and the angle A equal to the angle D. It is required to prove that the triangles are equal in every respect; that is, the bases, the areas, and the other angles, namely, those angles which are opposite to the equal sides. For if the triangle ABC be applied to DEF, so that the point A may be on D, and AB on DE, then the point B shall be on E, because AB is equal to DE. (Hyp.) And AB coinciding with DE, AC shall coincide with DF, since the angle A is equal to the angle D, (Hyp.) therefore also the point C shall be on F; because AC is equal to DF; (Hyp.) but the point B coincides with E, therefore the line BC shall lie wholly on EF; (I. Def. 5, cor.) hence the bases BC and EF are equal. (Ax. 8.) And since the triangles have been proved to coincide, the areas are equal; (Ax. 8.) Also, since the sides AB and BC coincide with DE and EF the angles B and E are equal; (Ax. 8.) and since AC and CB coincide with DF and FE, the angles C and F are equal; (Ax. 8.) and these are the angles opposite to the equal sides. Hence the triangles are equal in every respect. 2. (Euc. I., 26 Pt. I.). Next, if the angle B is equal to the angle E, the angle C equal to the angle F, and the side BC equal to the side EF. It is required to prove that the triangles are equal in every respect. For if the triangle ABC be applied to DEF, so that the point B shall be on E, and BC on EF, then the point C shall be on F, because BC is equal to EF. And BC coinciding with EF, BA shall coincide with ED and CA with FD, since the angle B is equal to the angle E, and the angle C equal to the angle F. (Hyp.) (Hyp.) (Hyp.) Then the point A, which is on BA and CA, must lie on ED and FD, that is, it must lie on the point D; hence the one triangle exactly coincides with the other, Schol.-If the sides AB and DE be produced to G and H, the angles below the base shall also be equal. For, since AB and DE coincide, their continuations BG and EH will also coincide (f, p. 16), so that GB and BC coincide with HE and EF; and therefore the angles GBC and HEF are equal (ax. 8); and the same would appear if AC and DF were produced. PROP. V.-THEOREM. The angles at the base of an isosceles triangle are equal, and if the equal sides be produced, the angles on the other side of the base are also equal. Let ABC be an isosceles triangle, having the side AB equal to the side AC; it is required to prove that the |