A Let ABC, DEF be two isosceles A°, having the sides AB, AC of the one to each other, and also to the sides DE, DF of the other. Also let the included L BAC = included LEDF; then the Ls B and c respectively LS E and F. (Prop. IV.) But since AB = DF, and AC = de, if the A DEF be B E turned over, and applied to the A ABC, so that the DF may coincide with AB, and be with AC, the L will coincide with 1 B, and the L E with c, as in Prop. IV.; .. the /s B and c are each = E, and .. they each other.-Q. E. D. In forming roofs, brackets, tripods, and in various operations of Carpentry, this principle is constantly applicable. с F PROP. VI. THEOR. A Gen. Enun.-If two angles of a triangle be equal to one another, the sides also which subtend, or are opposite to, the equal angles, shall be equal to one another. PART. ENUN.—Let ABC be a.A, of which the ABC = ACB, then the side AB = AC. CONST.–For if AB be not = AC, one must be > the other. Suppose AB to be the >, and that from it is cut off dB = AC, the <. (Prop. III.) Join DC (Post. 1). DEMONST.-1. Show the absurdity resulting from the above supposition ; viz., that A DBC would = A ABC. D .B с Since in AS DBC, ACB, the side dB = AC, and BC is common to both, ... the two DB, BC = the two AC, CB, each to each ; and the < DBC = Z ACB (Hyp.); ·:. base dc = base AB, and the area of A DBC = area of a ACB : i.e. the < = >, which is absurd. 2. Hence AB is not unequal to ac: i.e. it is = to it. Wherefore, if two Zs of a triangle, &c.— Q. E. D. CoR.—Hence every equiangular - is also equilateral. To prove this corollary, let the _s of the A ABC be all equal to one another : then because the L 2 ACB, .. AB = AC; and because the | ABC = AC; .'. AB i. e. the A ABC is equilateral.—Q. E. D. B The Sixth Proposition is proved by what is called a reductio ad absurdum ; that is, by assuming an opposite conclusion to that which is enunciated, and showing it to be inconsistent with the hypothesis. It is the converse of the former part of the fifth ; and though in this case, as in others, Euclid has only demonstrated what was necessary to the proof of subsequent propositions, the second part is also convertible. It is not, however, to be supposed that, in taking the conclusion for an hypothesis, the original hypothesis becomes a necessary conclusion. Instances will be noticed in which this is not the case. In the meantime, the converse of the latter part of the Fifth Pro. position may be thus demonstrated. ABC 2 = BC : BAC, .. BC = AC с PROP. C. THEOR. Gen. ENUN.-If, when two sides of a triangle are produced, the angles below the base are equal to each other, the two sides are also equal to one another. B с F D E Part. ENUN.—Let ABC be a A; and A the sides AB, AC being produced to D and E, let the 2 DBC = L ECB; then the side AB shall the side AC. Const.-In BD take any pt. F, and LCBG. Because BF = cg, and Bc is common to the AS FBC, GCB, .. the two sides FB, BC = GC, CB, each to each; and the contained / FBC GCB (Hyp.); .. the base FC base GB, and the / BCF 2 CBG. (Prop. IV.) 2. Prove that L GCF = LFBG. But / BCG 2 CBF (Hyp.); .. by subtraction, L GCF = L FBG (Ax. 5). 3. Prove that / BFG = LCGF. GC, CF, each to each ; and the / FBG = LGCF ; .. LBFG L CGF. (Prop. IV.) 4. Prove AF = AG. Now these are the at the base of the A AFG; ... AF = AG. (Prop. VI.) 5. Prove AB = AC. But BF = cg (Const.); ... by subtraction, AB = AC. Wherefore, if, when two sides of a A &c.-Q. E. D. PROP. VII. THEOR. GEN. Enun.—Upon the same base, and on the same side of it, there cannot be two triangles that have their sides, which are terminated in one extremity of the base, equal to one another, and likewise those which are terminated in the other extremity. D A B PART. ENUN.-Let the two AS ACB, ADB be upon the same base AB, and the same side of it, and let the sides CA, DA, terminated in A, be =; then the sides CB, DB, terminated in B, cannot be =. CONST.-Join c. DEMONST. (By reduct. ad absurdum.-If possible, suppose CB = DB. Case 1. When the vertex of each s is without the other A. 1. Prove that L. BDC is > 2 BCD. Because AC = AD (Hyp.), .. 2 ACD = _ ADC. (Prop. V.) But < ACD > BCD (Ax. 9); ... Z ADC > Z BCD. Much more is < BDC >. <BCD. 2. Prove that, also, on the above supposition, 2 BDC = 2 BCD. Now if BC = BD, .. 2 BDC = Z BCD. (Prop. V.) 3. But it has been shown that 2 BDC is > BCD, which is impossible. Case 2. When one of the vertices, as D, is within the other A ACB. Const.–Produce AC, AD to E and F, and join cd. DEMONST. (Ad abs.) D E F D A B < $ ECD, FDC, upon the other side of the base CD, are equal. (Prop. V.) But the Z ECD > Z BCD (Ax. 9); .. FDC > < BCD. Much more is Z BDC > Z BCD. 2. Also that they are =. Now, if in A BCD, BC = BD, .. 2 BDC = BCD. (Prop. V.) 3. But it has been shown that 2 BDC is > BCD, which is impossible. The third case, in which the vertex of one o is upon a side of the other, needs no demonstration. (See figure to Prop. VI.) Wherefore, upon the same base, and on the same side of it, &c.-Q. E. D. In the same way it may be proved that, if the sides BC, BD are equal, then AC, AD cannot be equal ; and the student will do well to work out the demonstration. The proof includes what is called a demonstration à fortiori. Euclid has established this theorem for the sole purpose of proving the next, to which it may therefore be considered as a Lemma ; i. e, a proposition which is merely preparatory to the demonstration of another. The limitation that the two triangles should be upon the same side of the base is manifestly essential; for upon A the other side of the base AB, a A ABD may be constructed in every respect similar and = to the A ABC. (Prop. IV.) с B |