PROP. VIII. THEOR. GEN. Enun.-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 which is 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. PART. ENUN.—Let ABC, DEF be two As, having the sides AB, AC = to the two sides DE, DF, each to each ; likewise let the base BC base EF; then < BAC shall be = LEDF. DEMONST.-(By supraposition). 1. For if the A ABC be applied to the a DEF, so that the pt. B be on E, and the st. line BC on EF, the pt. c shall coincide with F, be в с E E cause BC = EF. (Hyp.) 2. Now Bc coinciding with EF, AB, AC shall also coincide with DE, DF. For if AB, AC, do not coincide with DE, DF, but take another direction, as, for instance, GE, GF, then upon base upon the same side of it, there can be two AS DEF, GEF, which have their sides terminated in one extremity of the same EF, and F BCD. A B в < $ ECD, FDC, upon the other side of the base CD, are equal. (Prop. V.) But the Z ECD > ZBCD (Ax. 9); .. ZFDC > < Much more is 2 BDC > Z BCD. 2. Also that they are =. Now, if in A BCD, BC = BD, .. <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 sis upon 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.) a с B PROP. VIII. THEOR. G Gen. Enun.-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 which is 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. Part. ENUN.—Let ABC, DEF be two having the sides AB, AC = to the two sides DE, DF, each to each ; likewise let the base BC = base EF; then < BAC shall be = L EDF. DEMONST.-(By supraposition). 1. For if the A ABC be applied to the DEF, so that the pt. B be on E, and the st. line BC on EF, the pt. c shall coincide with F, because BC = EF. (Hyp.) 2. Now Bc coinciding with EF, AB, AC shall also coincide with DE, DF. For if AB, AC, do not coincide with DE, DF, but take another direction, as, for instance, GE, GF, then upon the same base and upon side of it, there can be two As DEF, GEF, which have their sides terminated in one extremity of B EF, the same the base = to one another, and also those which are terminated in the other extremity, which is impossible. (Prop. VII.) 3. If ... the base bc coincides with EF, the sides AB, AC must coincide with DE, DF; and .:. also the < BAC coincides with 2 EDF, and is = to it. Wherefore, if two As have two sides, &c.Q. E. D. This Proposition is the converse of the former part of the fourth. It contains the second condition of equality of two or more As; and the coincidence of the 4s contained by the two equal sides being proved, the equality of the areas and the remaining / 8 follows from the Fourth Proposition. PROP. IX. PROB. A Gen. Enun.-To bisect a given rectilineal angle; that is, to divide it into two equal angles. PART. ENUN.—Let Bac be the gn. rectilineal Z; then it is required to bisect it. Const.—In AB take any pt. D, and from AC cut off AE = AD. (Prop. III.) Join DE; and upon it describe an equilateral A DEF. (Prop. I.) Join AF. The st. line AF bisects the 2 BAC. DEMONST.-Because AD=AE, and AF is common to AS DAF, EAF, .. the two sides DA, AF = EA, AF, each to each; and the base DF = base EF (Const.); :: DAF = LEAF. (Prop. VIII.) B E line, the 2. DBE = LEBA (Def. 10); .. the 2 DBE = L CBE: i. e. the <= >, which is impossible. Therefore two straight lines cannot have a common segment.-Q. E. D. D A с B This corollary is placed by Euclid himself among the Axioms, and the simple proof, deduced from the Eleventh Proposition, was added by Dr. Simson. From the plain fact, however, that two straight lines cannot meet in more than one pt. without coinciding altogether, it is clear that a segment, or part cut (seco) from one straight line, cannot possibly belong to another also. With respect to the Proposition itself, it is clear that, if the gn. pt. be at or near the extremity of the line, it will be necessary to produce it: but if such production be impossible or inconvenient, the following method may be used : PROP. D. PROB. D E GEN. ENUN.-Through the extremity of a given finite straight line to draw a straight line at rt. sto it. Part. ENUN.—Let AB be a gn. finite st. line; then it is required to draw through b a st. line at rt. $ to AB. CONST.-In AB take any pt. c, and through c draw CD at rt. 28 to AB (Prop. XI.), and make (Prop. III.) Bisect the LBCD in the A st. line ce. (Prop. IX.) Again, through D draw de at rt. 2to cd (Prop. XI.), meeting ce in E. Join Be (Post. 1). Then be will be at rt. 4s to AB. DEMONST.-Because cb = cd, and CE is common to the A8 ECB, ECD; .. the two sides BC, CE = DC, CE, each to each, and the LBCE = L DCE; ... the L CBE = L CD св. .B |