The converse of the Proposition may be thus demonstrated: 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 C 25 CEA, DEB = to one another; then CE shall be in the same st. line with DE. DEMONST.-1. Because the ▲ CEA = = DEB (Hyp.), add to each the A S E B CEB; .'. < CEA + CEB = < CEB and DEb (Ax. 2). 2. But the CEA + CEB = two rt. 4 (Prop. XIII.); .. the CEB + DEB = two rt. 4s (Ax. 1). 3. .. CE is in the same st. line with DE. (Prop. XIV.) 4. In the same manner, if the CEB = AED, 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 PROP. H. PROB. 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. C CONST.-From c draw CE LA- E 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. D -B G DEMONST.-Because CE = FE, and EG is common to the ▲ CEG, FEG; .. the two sides CE, EG = FE, EG, each CEG = rt. FEG; .'. the CGE to each, and the rt. = FGE (Prop. IV.) = vertical / DGB (Prop. XV.): i. e. the two st. lines CG, DG, drawn from C, D, and meeting 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 ▲ ABC be produced to D; then the exterior ACD is > than either of the exterior and opposite CBA, BAC. s 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 ▲ EAB, ECF, the BAE = ▲ ECF. Because AE EC, and BE = EF (Const.); .. the two sides AE, EB = CE, EF, each to each, and the vertical AEB = < CEF (Prop. XV.); .. the base AB = base CF, and the AEB A CEF, and the remaining 4s of the one = the remaining s of the other, to which the B K A E H C F G equal sides are opposite; .. the BAE = ≤ ECF. (Prop. IV.) 2. But the ACD is > ECF; ACD is BAC. the 3. In the same manner it may be shown, by producing AC to G, and bisecting BC, that the BCG, which is equal to the vertically opposite ACD (Prop. XV.), is > than the 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 к, making нK = AH; and join KC. DEMONST.-1. Then, because BH = HC, and ah = hk in the ▲ AHB, CHк; .. the two sides BH, HA = сн, нк, each to each, and the vertical / BHA Z CHK (Prop. XV.); .. the ▲ ABH = LHCK. = 2. But the BCG is > нCK; .. the ▲ BCG is > ABC. 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 ▲ 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 ▲ ABC be moved along the line BE till в 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 ACE. In other words, the interior ABC F E will have moved into the position of the DCE, which is <the ACE. In like manner, if the ▲ be moved along the line AF, the A will be seen to be less than the BCF, which = ACE. This Proposition is one of those which are not convertible; for it does not follow that, if the exterior of a rectilineal figure be> than its interior opposite 4s, the figure is necessary a A. This will be evident from the following Theorem : PROP. K. THEOR. 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. E PART. ENUN.-Let ABCD be a trapezium, of which the Z BAD is obtuse, and the ADC a rt. ; 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 / BAD, ADC, DCB. DEMONST.-1. For the exterior LEBG of A ABE, is > interior EAB. (Prop. XVI.) D C G .., a fortiori, CBG is the rt. ADC (Def. 11). 3. Again, in the ▲ CFB, the exterior terior FCB. (Prop. XVI.) 4. .. the exterior BAD, ADC, DCB. CBG is > in CBG is > either of the interior s 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 4o, formed by producing CB towards B, and the sides AB and AC both ways. By way of deduction may be proved, PROP. L. THEOR. GEN. ENUN.-Only two equal straight lines can be drawn to another straight line from a given point without it. PART. ENUN.-Let AB be a st. line; c a 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 CB. DEMONST. (Ad abs.) If possible, suppose CD = CA 1. Now, since CA = CB,.. cause CA = CD, .'. ▲ CAB = 2 CBA (AX. 1). CDA = = CB. CAB = CBA; and be- CDA is > interior 2. But in the ▲ CDB, exterior CBA (Prop. XVI.), being also to it, which is absurd. 3... CD is not to CA or CB. = to them. 4. In the same manner it may be shown that no other line drawn from the gn. pt. c to AB is Wherefore, only two &c.-Q. E. D. = COR.-A circle cannot cut a st. line in more pts. than two. PROP. XVII. THEOR. GEN. ENUN.-Any two angles of a triangle are together less than two right angles. PART. ENUN.-Let ABC be any A. Then any two of its Zare together less than two rt. s. DEMONST.-Produce BC to D (Post. 2). 1. Now the exterior is interior opposite To each of these add the B A C D ACD of the ▲ ABC SACD + ACB are > the (Ax. 4). S ABC + ACB 2. But the < ACD + ACB = two rt. s |