The converse of the Proposition may be thus demonstrated : А E B PROP. G. THEOR. Gen. Enun.-If at a point in a straight line, two other straight lines, upon the opposite sides of it, make the vertical angles equal, these two straight lines shall be in one and the same straight line. Part. ENUN.-At the pt. E, in the st. line AB, let the two st. lines CE, DE make the vertical * CEA, DEB = to one another; then CE shall be in the same st. line with DE. DEMONST.-1. Because the L CEA = = DEB (Hyp.), add to each the L CEB; .; Z S CEA + CEB = LS CEB and DEB (Ax. 2). 2. But the 4 CEA + CEB = two rt. 48 (Prop. XIII.); ... the L CEB + DEB two rt. 48 (Ax. I). 3. CE is in the same st. line with DE. (Prop. XIV.) 4. In the same manner, if the 2 CEB = LAED, it may be shown that ce and dE are in the same st. line. Wherefore, if at a pt. &c.-Q. E. D. As a deduction from the 15th Proposition may be proved D D Gen. Enun.-From two given points on the same side of a straight line given in position, to draw two straight lines which shall meet the given line in the same point, and make equal angles with it. Part. ENUN.-Let AB be the gn. st. line, and c, d two gn. pts. on the same side of it; then it is required, &c. CONST.-From c draw CEI A to AB (Prop. XII.); produce CE to F (Post. 2), and make EF = CE (Prop. III.); join DF, CG (Post. 1); then cg, dg will be the lines required. E B G DEMONST.-Because CE = FE, and EG is common to the ASCEG, FEG; .. the two sides CE, EG = FE, EG, each to each, and the rt. L CEG = rt. / FEG; .. the / CGE | FGE (Prop. IV.) vertical / DGB (Prop. XV.): i. e. the two st. lines cG, DG, drawn from C, D, and meeting in G, make s with AB.-Q. E. F. PROP. XVI. THEOR. GEN. ENUN.-If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles. Part. Enun.—Let a side bc of any A ABC be produced to D; then the exterior 2 ACD is >than either of the exterior and opposite 2$ CBA, BAC. Const.—Bisect ac in E (Prop. X.); join BE, and produce it to F, making EF = BE (Prop. III.); and join Fc. DEMONST.-1. Prove that in the AS EAB, ECF, the Z BAE = ECF. Because AE = Ec, and BE = EF (Const.); ... the two sides AE, EB = CE, EF, each to each, and the vertical Z AEB = - Z CEF (Prop. XV.); ... the base AB = base CF, and the A AEB = the remaining zs of the one = the remaining zs of the other, to which the equal sides are opposite; .. the Z BAE = 2 ECF. (Prop. IV.) A F E E H н IC B D O CEF, and 2. But the < ACD is > Z ECF; .. the ACD is > Z BAC. 3. In the same manner it may be shown, by producing ac to G, and bisecting BC, that the 2 BCG, which is equal to the vertically opposite 2 ACD (Prop. XV.), is > than the 2 ABC. Wherefore, if one side of a be produced, &c.-Q. E. D. The second part of the Proposition is proved thus : CONST.-Produce ac to g; bisect bc in h; join Ah, and produce it to k, making hk = Ah; and join kc. DEMONST.-1. Then, because BH = HC, and au = in the AS AHB, CAK; : . the two sides BH, HA = ch, HK, each to each, and the vertical / BHA CAK (Prop. XV.); .. the ABH = LHCK. 2. But the BCG is > | HCK; ... the / BCG is > HK ABC. D 3. .. also the vertical / ACD is > ABC.-Q. E. D. It is observable that the exterior / ACD will be either >, =, or < than the adjacent , ACB, according as the A ABC is acute-angled, right-angled, or obtuse-angled respectively. The following is a practical illustration of the property set forth in the Proposition : Let the A ABC be moved along the line be till B coincides with c. B The vertex of the A will now be to the right of A, so that dc will be within the exterior In other words, the interior , will have moved into the position of the 2 DCE, which is < the LACE. In like manner, if the A be moved along the line AF, the L A will be seen to be less than the / BCF, which 2 ACE. ABC F ACE. This Proposition is one of those which are not con. vertible ; for it does not follow that, if the exterior of a rectilineal figure be > than its interior opposite 2$, the figure is necessary a A. This will be evident from the following Theorem : PROP. K. THEOR. E D с Gen. Enun.-If a trapezium, having two of its adjoining angles an obtuse and a right angle respectively, be such that its opposite sides, produced in the direction of these angles, shall meet, the exterior angle in the other direction shall be greater than any of the three interior opposite angles. Part. Enun.—Let abcd be a trapezium, of which the | BAD is obtuse, and the L ADC a rt. L; and let the sides AD, Bc be produced to meet in E, and BA, CD in F; then, if AB be produced towards G, the exterior / CBG shall be > than either of the three / S BAD, ADC, DCB. DEMONST.-1. For the exterior 2 EBG of A ABE, is > interior / EAB. (Prop. XVI.) 2. But | EAB is an obtuse l ; .., a fortiori, L CBG is > the rt. L ADC (Def. 11). 3. Again, in the ACFB, the exterior LcBG is > interior / FCB. (Prop. XVI.) 4. .. the exterior LcBG is > either of the interior / s BAD, ADC, DCB. Wherefore, if a trapezium, &c.-Q. E. D. The student cannot be too frequently reminded that the Propositions of Euclid are of universal application, and in this 16th Proposition he will find it a useful exercise to prove its truth with respect to the other exterior 28, formed by producing cb towards B, and the sides AB and AC both ways. By way of deduction may be proved, G F A B PROP. L. THEOR. Gen. Enun.- Only two equal straight lines can be drawn to another straight line from a given point without it. A D B PART. ENUN.--Let AB be a st. line ; ca gn. pt. without it; and ca, CB two = st. lines drawn from c to AB; then no other st. line can be drawn from c to AB, which shall be to CA or B. DEMONST. (Ad abs.) 1. Now, since CA = CB, .. / CAB = L CBA; and because CA = CD, .. 2 CAB = CDA (Prop. V.); ..L CDA CBA (Ax. 1). 2. But in the A CDB, exterior / CDA is > interior L CBA (Prop. XVI.), being also to it, which is absurd. 3. .. cp is not to CA or CB, 4. In the same manner it may be shown that no other line drawn from the gn. pt. c to AB is = to them. Wherefore, only two &c.-Q. E. D. COR.-A circle cannot cut a st. line in more pts. than two. PROP. XVII. THEOR. A GEN. ENUN.-- Any two angles of a triangle are together less than two right angles. PART. ENUN.-Let ABC be any Δ. Then any two of its <s are together less than two rt. Zs. DEMONST.-Produce Bc to D (Post. 2). 1. Now the exterior 2 ACD of the A ABC is > interior opposite 2ABC. (Prop. XVI.) To each of these add the < ACB; .. the ZS ACD + ACB are > the ZS ABC + ACB (Ax. 4). 2. But the ZS ACD + ACB = two rt. 28 B с D |