20 Th. Any one side of a triangle is less than the su of the other two. Constr. In the triangle ABC, produce BA so as to make AD equal to AC: join CD (a). Argument. The angles ACD, ADC are equal, being opposite to equal sides (a); but either of them is less than BCD; and the less side subtends the less angle (6); therefore BC is less than BD, which is the sum of BA and B In this manner it may be proved that either side is less than the sum of the other two. Q. E. D. Recite (a), pos. 1; (b), p. 5, 19. AC. 21 Th. If from a point (D) within a triangle (ABC), two straight lines (DB, DC) be drawn to the ends of the base; these lines shall be less than the two sides (AB, AC), but they shall contain a greater angle. Argument 1. Produce BD to E (6). BE is less than the sum of BA, AE (6); add EC to both. Then the sum of BE and EC is less than the sum of BA and AC (c). Again CD is less than the sum of CE and ED (6); to both add DB; then the sum of CD, DB is less than the sum of CE, EB, and still less than the sum of CA, AB (c). 2. The angle BEC is exterior of the triangle BAE, and interior of the triangle CED; it is therefore less than the exterior BDC, and greater than the interior BAC : BDC is therefore greater than BAC (d). Wherefore, if from a point, &c. Q. E. D. Recite (a), pos. 2; (6), p. 20; (c), ax. 4; (d), p. 16. B В 22 P. To make a triangle of which the sides shall be equal to three given straight lines (A, B, C); but any one of them must be less than the sum of the other two. Constr. Take a straight line DE, not less than the sum of the three given straight lines; make DF equal to A, FG equal to B, and GH equal to C (a): from F as centre, and radius FD, describe the circle DKL (6); also from G, as centre, and radius GH, describe the circle KHL. The triangle FGK has its sides equal to the straight lines A, B, C. А Argument. FK, FD are equal radii (c); and FD equals A; therefore FK equals A (d): also GK, GH are equal radii; and GA equals C; therefore GK equals C. Therefore FK, FG, GK are equal to A, B, C, each to each; and the thing is done which was required. Recite (a), p. 3; (b), pos. 3; (c), def. 15; (d), ax. 1. 23 P. At a given point (A) in a given straight line (AB) to make an angle equal to å given rectilineal an. gle (C). Constr. In the lines forming the angle C join any two points D, E (a); make the triangle AFG of sides equal to CD, CE, DE, each to each (b). Argument. Since FA, AG are made equal to DC, CE, and FG to DE; therefore, at the point A, in the straight line AB, an angle is made equal to the angle C (c); which was to D be done. Recite (a), pos. 1; (b), p. 22; (c) p. 8. F 24 Th. If two sides (AB, AC) of one triangle, be equal to two sides (DE, DF) of another, while the contained angles (A and D) are unequal ; the base (EF) oppo. site the less angle (D) is less than the base (BC) opposite the greater angle (A). Constr. Let DF be not less than DE: then, at the point D, in the straight line ED, make the angle EDG equal to A (a); make DG equal to AC, or DF (V); join EG, GF (c). Dem. The two sides BA, AC being severally equal to the two ED, DG, and the angle A to EDG, the bases BC, EG are therefore equal (d). But, in the triangle EFG, EF is less than EG; for it subtends a less angle--that is, since DFG is isosceles (d), the angles DFG, DGF are equal; but EGF is less, and EFG is greater than either : therefore EF is less than BC. Wherefore, if two sides, &c. Q. E. D. Recite (a), p. 23; (6), cor. p. 3; (c), post. 1; (d), p. 5. AA 25 Th. If two sides of one triangle be severally equal to two sides of another, but the bases unequal; the angle opposite the greater base is greater than the other. Dem. If, in the triangles ABC, Wherefore, if two sides, &c. B Q. E. D. 26 Th. Two triangles are equal to one another, in all their parts, which have two angles and a side in the one equal to two angles and a side in the other; whether the equal sides be adjacent to both, or only to one of the equal angles. А B с E Let ABC, DEF be two triangles, having equal angles at B, E and at C, F. 1. Let BC=EF, which are adjacent to both the equal angles; then, the side AB, for example, will equal the sile DE: but if not equal, let it be greater; so that a part of it, as BG, shall equal ED (a) join GC;-Hence the triangles BCG, EFD are equal, having the angles B, E equal, and the sides BC, BG FB in the one equal to the sides EF, ED in the other (6). Therefore, the angles BCG and F are equal ; but the angles BCA and F are equal, by hypoth. therefore BCG a part, equals BCA the whole, which cannot be admitted. Therefore, AB is not greater than DE. 2. Let AB=DE, which are adjacent to the equals B, E, and opposite to C, F: then, the sides BC, EF, for example, are equal; but if not equal, let BC be the greater; so that a part of it, as BH, shall equal EF (a). Hence the triangles ABH, DEF are equal, having he angles B, E eqnal by hypoth. and the sides BA, BH equal to he sides ED, EF (6): therefore the angle BHA equals the angle F, or its equal C; that is, the exterior equals the interior on the same side (), which is impossible. Therefore BC is not greater than EF. Wherefore, two triangles are equal in all their parts, &c. Q. E. D. Recite (a), p. 3; (b), p. 4; (c), p. 16. 27 Th. If a straight line (EF) fall upon two other straight lines (AB, CD), making the alternate angles (AEF, EFD), equal to each other; these two straight lines shall be parallel Argument. For, if AB be not saral« 1 to CD, produce them, and they shall diverge in one direction and meet in the other. Let them meet in the point G: therefore GEF is a triangle, whose exterior angle AEF exceeds its interior opposite angle EFG (a), which were said to be equal. Therefore AB and CD, neither meet nor diverge, by production, as was supposed; but are parallel, (6). Q. E. D. Recite (a), p. 16; (6), Note def. 4. с F 28 Th. If a straight line (EF), falling upon other two straight lines (AB, CD), make equal the exterior angle (EGB) to the interior opposite angle (GHD on the same side of the line ; or make the interior angles (BGH, GHD) equal to two right angles, the two lines shall be parallel. Argument 1. Since, by hyp. EGB is equal to GHD, and also to AGH (a); therefore AGH is equal to GHD (b); and they are alternate angles (c); therefore AB is parallel to CD. 2. Since by hyp. BGH and GHD are eqnal to two right angles, and that AGH and BGH are also equal to two right angles (d), the former two are equal to the latter two (6); then taking out the common angle BGH, the remain AGH equal to GHD (e); and they are alternate angles (c): therefore AB is parallel to CD. Wherefore, if a straight line, &c. Q. E. D. Recite (a), p. 15; (6), ax. 1; (c), p. 27; (d), p. 13; (e), ax. 3. 29 Th. If a straight line (EF) cut two parallel straight lines (AB, CD), it makes equal 1, the alternate angles ; 2, the interior and exterior angles on the same side; and 3, the two interior angles on the same side to two right angles. 1. The alternate angles AGH, GHD are equal, or not; if not. let AGH be the greater, and add BGH to each ; then AGH and BGH exceed BGH and GHD (a): but AGH and BGH are equal to two right angles (6); therefore BGH and GHD are less than two right angles, and the lines AB, CD will meet towards B, D (c); but they are parallel and cannot meet; therefore the alternate angles AGH, GHD are equal, as stated. H н 2. The exterior and interior angles EGB, GHD, on the same side of EF are equal : fur, as vertical angles EGB, AGH are equal A(d); and AGH proves equal to GHD: therefore EGB equals GHD, as stated. 3. The interior angles BGH, GHD are equal to two right angles : for AGH and BĜH are equal to two right angles (6); and AGH equals GHD: therefore BGH and GHD are equal to the same, as stated. Wherefore, if a straight line, &c. (6) p. 13; Q. E. D. 2 B E 30 Th. Straight lines (AB, CD) which are parallel to the same straight line (EF), are parallel to each other. Let GHK cut the parallels AB, EF, also CD, EF; then AB is parallel to CD. For since the parallels AB, EF are cut by a straight line, the alternate angles AGH, A GHF are equal (a): Also, because the parallels CD, EF are cnt by a straight line, the exterior angle CGHF equals the interior, opposite angle GKD (6) Therefore, each of the angles AGH, GKD, being equal to GHF, are equal to each other (c); and being alternate angles, AB is parallel to CD (a). Wherefore, straight lines, &c. Q. E. D. Recite (a) p. 27 ; (b) p. 28; (c) ax. 1. 31 P. To draw a straight line through a given point (A), parallel to a given straight line (BC). D In BC take any point D, and join DA (a); make the angle DAE equal to the angle ADC (b); produce the straight line EA to F (c). Because the alternate angles EAD and ADC are equal (d), the straight line EF is B parallel to the given straight line BC; and it is drawn through the given point A, which was to be done. Recite (a) pos. 1 ; (b) p. 23; (c) pos. 2; (d) p. 27. |