The 8th axiom is often expressed thus :—“Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another.” But the explanatory clause in italics cannot apply to angles and straight lines, which do not fill space. Unless Axiom 10 were assumed, there would be a difficulty about applying Axiom 8 to rectilineal figures. For, to begin with a straight line, when we have fitted the two ends of one straight line upon the two ends of another, we could not be sure that the lines are entirely fitted one upon the other, if it were possible for them to enclose a space. We shall find excellent practice in the applications of Axioms 8—10 in the Exercise preliminary to Proposition IV. 21. PARALLELISM. In order to complete the collection of axioms, the 12th will now be given. It will be much better, however, not to discuss it until we have made some further progress in the subject. 12. If a straight line meet two straight lines so as to make the two interior angles on the same side of it taken together less thantworight angles, these straight lines, being continually produced, shall at length meet on that side on which are the angles which are less than two right angles. EXAMINATION XVI. (Recapitulatory.) 1. What sort of facts are called axioms? 2. If a fact be very easy to believe without proof, but at the same time is capable of proof, would Euclid admit it as an axiom? 3. The first axiom speaks of “things'; does this necessarily refer to things occurring in Geometry only? 4. Explain the words “bisect'; and say what is meant when a magnitude is said to be common to two others. 5. State the meaning of superposition.' 6. To what use does Euclid put the process of superposition ? 7. State another form of Axiom 8 which is often used, and show its defect. 8. What axiom may be used to show that the lesser of two given straight lines may be made to lie along a part of the greater ? THEOREMS AND PROBLEMS. 22. WHAT ARE PROPOSITIONS AND PROBLEMS? 6 Euclid's reasoning, in his “Elements,” is divided into numerous portions, generally called 'Propositions.' Strictly speaking, a Proposition is that which places before us the object of some piece of reasoning upon which we are about to enter. The word comes from the Latin pro, before, and pono, I place. But, just as a circle is defined to be one thing and often used for another, so the meaning of Proposition in Geometry is very commonly extended, so as to embrace the whole argument to which it belongs. This extension of meaning brings into use another word, namely • Enunciation. This is from the Latin enuntio, I announce or declare, and, in general, simply means a statement or declaration ; while, in Euclid, it may be defined as follows: The Enunciation of a Proposition is the general statement of its object, or the purpose with which it commences. There are two classes of Propositions, called Problems and Theorems. We shall meet first with Problems. A Problem is a Proposition which proposes that some geometrical construction shall be effected. The Enunciation of a Problem is divided into two parts, namely, the datum (or, plural, data), and the quaesitum (or, plural, quaesita). In Latin, datum means something given, and quaesitum something sought. Hence The data of a problem are the things which are given. The three postulates (Art. 15) are sometimes spoken of as 'self-evident' problems. Solving'a problem means, finding out how to do what its enunciation requires us to do. In theoretical Geometry (such as Euclid's), after stating the solution of a problem, we usually add a proof that the solution is a correct one. It is to be remembered, in solving our problems, that we are restricted to a straight-edge and compasses to work with. A ‘finite' straight line is limited in length, having two extremities given or understood. D EXERCISE III. Distinguish in each of the following little problems 1–6 what is given from what is required; then give a solution and proof. A full answer to No. 1 is given, which should be followed as a model in working Nos. 2–6, all definitions and postulates being carefully quoted in the margin, as shown. 1. To produce a finite straight line until it becomes twice as long as at first. An answer to this might be given as follows : Given : a finite straight line. Solution. Taking B as centre, and BA as radius, draw the circle ADC; (Post. 3) and produce AB until it meets the circle again in the point C. Proof (Def. 15) 2. Given a circle, its centre, and a point in its circumference, to draw through this point a diameter. 3. Given two points, draw a circle which shall have one point as centre and the other in its circumference. 4. Given a circle and its centre, make an isosceles triangle having its vertex at the centre. 5. Given a finite straight line, to make an isosceles triangle with this line for one of its equal sides. (Use Post. 3 first.) 6. Given a finite straight line and its middle point, to draw a semicircle B a a upon it. a 7. Take two points A, B and join them; (Post. 1) with centre A and radius A B describe a circle ; (Post. 3) and with centre B and radius B A describe another circle ; produce AB to meet one circumference in C and BA to meet the other in D. (Post. 2) In the figure thus constructed, how many times is AB contained in CD ? By what authority may you assume BC equal to BA, and also AD equal to AB? 8. By what axiom may we infer that BC and AD are therefore equal ? SECTION I. PROPOSITIONS I.-VI. First case of Equality of Two Triangles, with accessory Problems. 23. PROPOSITION 1.- PROBLEM. To describe an equilateral triangle upon a given finite straight line. Let AB be the given finite straight line. "It is required to describe an equilateral triangle upon AB. From the centre A, at the distance AB, describe the circle BCD; (Post. 3) and from centre B, at the distance BA, describe the circle ACE, cutting the former circle in C; and join CA, CB. Then ABC shall be an equilateral triangle. Because A is the centre of the circle BCD, therefore AC is equal to AB; (Def. 15) and because B is the centre of the circle ACE, therefore BC is equal to BA; wherefore AC and BC are each equal to AB. But things which are equal to the same thing are equal to one another; (Ax. 1) therefore AC is equal to BC; hence AB, BC, CA are equal to one another. Therefore the triangle ABC is equilateral ; (Def. 24) and it is described upon the given finite straight line AB. Which was to be done. Note.In the Exercises, where figures are not given although referred to, the student is required to draw them from the given directions. EXERCISE IV. 1. The circles of the foregoing figure intersect in another point besides C. Put F at this other point, and then prove, after the manner of Proposition I., that ABF is an equilateral triangle. 2. Prove also that ACBF will be a rhombus. 3. Take a straight line inclined like AB, B and construct an equilateral triangle upon it, on either side of it. 4. Also two equilateral triangles upon a straight line inclined like CD, one on each side of it. A 5. In the diagram of Proposition I. produce AB both ways to meet the circumference in D and E. With centre A and radius AE describe a circle; also with centre B and radius BD describe a circle : let these circles intersect on H and join H to A and B. Then show that HAB is an isosceles triangle, stating the magnitude of each side as compared with its base. 24. THE CONSTRUCTION OF A PROPOSITION. The Construction of a proposition describes the steps by which we complete the diagram so as to make it sufficient for the purposes of our reasoning. It usually precedes the proof, but may be sometimes found partially or wholly interspersed among the steps of the reasoning. When we wish to refer to the construction as authority for some subsequent statement, we place the abbreviation “ Constr." in the margin. Obs.-The letters R. E. F. are the initials of the Latin “ Quod erat faciendum,” and mean “which was to be done.” 25. PROPOSITION II. PROBLEM. From a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the given line. (Post. 1) upon AB describe the equilateral triangle DAB, (I. 1) |