Theorem. If two triangles have two sides of the one equal to two sides of the other, each to each; and have also the angles contained by those sides equal; the bases or third sides of the triangles shall be equal; as B E also the triangles; and their other angles shall be equal, each to each, viz., those to which the equal sides are opposite. Steps of the Demonstration. Conceiving the As applied to each other as directed, The 1st thing to be proved is— Note. The student who finds any difficulty in understanding this Theorem, which is the very foundation of the science of Geometry, may assist his apprehension of it, by cutting out the two triangles in paper, and actually applying one to the other according to the directions in Euclid. PROPOSITION V. Theorem. The angles at the base of an isosceles triangle are equal; and if the equal sides be produced, the angles upon the other side of the base shall be equal. Steps of the Demonstration. base BG base CF ABGACF of ▲ AFC, AGB AFC 1. Prove that in A GB Theorem. If two angles of a triangle be equal to one another, the sides also, which subtend, or are opposite to, the equal angles, are equal to one another. Note. The nature of the proof of this proposition is different from that of the preceding ones. This is an instance of the argument ad absurdum, i. e., it is proved that the theorem cannot be supposed false, without leading us to a manifest absurdity and contradiction, and we therefore conclude that it is true. And here let it be remarked that the usual editions of Euclid are calculated to mislead, or at least to puzzle, the learner, in their manner of proving these indirect propositions. For instance, instead of saying (as in Simpson), let AB be greater than AC, and from AB cut off DB = AC; and then referring the reader to Prob. III., to show him how to cut off this line, which, in fact, is not possible to be done, it would be B much more intelligible to say, Suppose AB to be greater than AC,-and suppose that a part of AB as DB AC, &c., We shall therefore, in giving the steps of the argument in such propositions, use this more correct mode of expression. Steps of the Demonstration. Having made the supposition that AB AC, but that one of them, as AB, is the other,-prove, on that supposition, 1. that A DBC = ▲ ACB, i.e. the less = greater, which shows the supposition to be false. 2. that.. AB is not ac, i. e., that Âв = 1c. PROPOSITION VII. (Argument ad Absurdum.) Theorem. On the same base, and on the same side of it, there cannot be two triangles that have their sides terminated in one extremity of the base equal to each other, and likewise those terminated in the other extremity. A B B Suppose it possible that the two AS ACB, ADB, on the same base AB can have the side CA of the one = side DA of the other, &c. (See Euclid.) Steps of the Demonstration in Case 1st, In which each of the vertices of the As falls without the other A. Prove on the above supposition,— BDC < ADC BCD (which shows that the 3. that BDC also = BCD (supposition is false. Steps of the Demonstration in Case 2nd, In which the vertex of one ▲ falls within the other A. Prove on the above supposition, PROPOSITION VIII. Theorem. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal, the angle contained by the two sides of the one shall be equal to the angle contained by the two sides, equal to them, of the other. Steps of the Demonstration. Conceiving the ▲ s applied to each other as directed, 1. Prove that the point c coincides with F, 2. 3. that BA, AC coincide with ED, df, (For we cannot suppose them not to coincide without falling into a contradiction,) that. BACEDF. Problem*. To bisect a given rectilineal angle; that is, to divide it into two equal angles. * The proofs of Propositions IX. XI. and XII. depend wholly on Proposition VIII. and that of X. wholly on Proposition IV. |