Fig. 23. Fig. 24. Fig. 25. DEF, coincide, the one with the other, and are equal in all respects. 39. Corollary. When, in two triangles, these three things are equal, namely, BC = EF, B = E, and C = F, we may thence infer, that the other three are also equal, namely, AB = DE, AC DF, and A= =D. THEOREM. 40. One side of a triangle is less than the sum of the other two. Demonstration. The straight line BC (fig. 23), for example, is the shortest way from B to C (3); BC therefore is less than BA+AC. THEOREM. 41. If, from a point O (fig. 24), within a triangle ABC, there be drawn straight lines OB, OC, to the extremities of BC, one of its sides, the sum of these lines will be less than that of AB, AC, the two other sides. Demonstration. Let BO be produced till it meet the side AC in D; the straight line OC is less than OD + DC; to each of these add BO, and BO+ OC<BO+OD+DC; that is BO+OC<BD+DC., Again, BD <BA+AD; to each of these add DC, and we shall have BD+DC<BA+AC. But it has just been shown, that BO+OC<BD+ DC; much more, then, is BO+ OC <BA + AC. THEOREM. 42. If two sides AB, AC (fig. 25), of a triangle ABC, are equal to two sides DE, DF, of another triangle DEF, each to each; if, at the same time, the angle BAC, contained by the former, is greater than the angle EDF, contained by the latter; the third side BC of the first triangle will be greater than the third side EF of the second. Demonstration. Make the angle CAGD, take AG = DE, and join CG; then the triangle GAC is equal to the triangle EDF (36), and therefore CG = EF. Now there may be three cases, according as the point G falls without the triangle ABC, on the side BC, or within the triangle. Case 1. Because GC<GI+1C, and AB<AI+IB, therefore GC+AB<GI+AI + IC + IB, that is, GC+ABAG + BC. From one of these take away AB, and from the other its equal Case II. If the point G (fig. 26) fall upon the side BC, then Fig. 26 it is evident that GC, or its equal EF, is less than BC. Case III. If the point G (fig. 27) fall within the triangle Fig. 27. BAC, then AG +GC<AB+BC (41); therefore, taking away the equal quantities, AG, AB, we shall have GCBC, or EFBC. THEOREM. 43. Two triangles are equal, when the three sides of the one are equal to the three sides of the other, each to each. Demonstration. Let the side AB = DE (fig. 23), AC = DF, Fig. 23. BC=EF; then the angles will be equal, namely, A = D, BE, and CF. For, if the angle A were greater than the angle D, as the sides AB, AC, are equal to the sides DE, DF, each to each, the side BC would be greater than EF (42); and if the angle A were less than the angle D, then the side BC would be less than EF; but BC is equal to EF; therefore the angle A can neither be greater nor less than the angle D; that is, it is equal to it. In the same manner it may be proved, that the angle B = E, and that the angle CF. 44. Scholium. It may be remarked, that equal angles are opposite to equal sides; thus, the equal angles A and D are opposite to the equal sides BC and EF. THEOREM. 45. In an isosceles triangle, the angles opposite to the equal sides are equal. Demonstration. Let the side AB= AC (fig. 28), then will Fig. 28. the angle C be equal to B. Draw the straight line AD from the vertex A to the point D, the middle of the base BC; the two triangles ABD, ADC, will have the three sides of the one equal to the three sides of the other, each to each, namely, AD common to both, AB= =AC, by hypothesis, and BD=DC, by construction; therefore (43) the angle B is equal to the angle C. 46. Corollary. An equilateral triangle is also equiangular; that is, it has its angles equal. Fig. 29. 47. Scholium. From the equality of the triangles ABD, ACD, it follows, that the angle BAD DAC, and that the angle BDA ADC; therefore these two last are right angles. Hence a straight line, drawn from the vertex of an isosceles triangle to the middle of the base, is perpendicular to that base, and divides the vertical angle into two equal parts. In a triangle that is not isosceles, any one of its sides may be taken indifferently for a base; and then its vertex is that of the opposite angle. In an isosceles triangle, the base is that side which is not equal to one of the others. THEOREM. 48. Reciprocally, if two angles of a triangle are equal, the opposite sides are equal, and the triangle is isosceles. Demonstration. Let the angle ABC ACB (fig. 29), the side AC will be equal to the side AB. For, if these sides are not equal, let AB be the greater. Take BDAC, and join DC. The angle DBC is, by hypothesis, equal to ACB, and the two sides DB, BC, are equal to the two sides AC, CB, each to each; therefore the triangle DBC is equal to the triangle ACB (36); but a part cannot be equal to the whole; therefore the sides AB, AC, cannot be unequal; that is, they are equal, and the triangle is isosceles. Fig 30. 49. Of the two sides of a triangle, that is the greater, which is opposite to the greater angle; and conversely, of the two angles of a triangle, that is the greater, which is opposite to the greater side. Demonstration. 1. Let the angle C>B (fig. 30), then will the side AB, opposite to the angle C, be greater than the side AC, opposite to the angle B. Draw CD, making the angle BCD = B. In the triangle BDC, BD is equal to DC (48); but AD+DC>AC, and AD+DC=AD + DB = AB; therefore AB>AC. 2. Let the side AB>AC, then will the angle C, opposite to the side AB, be greater than the angle B, opposite to the side AC. For, if C were less than B, then, according to what has just been demonstrated, we should have ABAC, which is contrary to the hypothesis; and if C were equal to B, then it would follow, that AC = AB (48), which is also contrary to the kypothesis; whence the angle C can be neither less than B, nor equal to it; it is therefore greater. THEOREM. 50. From a given point A (fig. 31), without a straight line Fig. 3k, DE, only one perpendicular can be drawn to that line Demonstration. If it be possible, let there be two AB and AC; produce one of them AB, so that BF AB, and join CF. = = : The triangle CBF is equal to the triangle ABC. For the angle CBF is a right angle (29), as well as CBA, and the side BF BA; therefore the triangles are equal (36), and the angle BCF-BCA. But BCA is, by hypothesis, a right angle; therefore BCF is also a right angle. But, if the adjacent angles BCA, BCF, are together equal to two right angles, ACF must be a straight line (33); and hence it would follow, that two straight lines ACF, ABF, might be drawn between the same two points A and F, which is impossible (25); it is, then, equally impossible to draw two perpendiculars from the same point to the same straight line. 51. Scholium. Through the same point C (fig. 17), in the Fig. 17. line AB, it is also impossible to draw two perpendiculars to that line; for, if CD and CE were these two perpendiculars, the angle DCB would be a right angle as well as BCE; and a part would be equal to the whole. THEOREM. 52. If, from a point A (fig. 31), without a straight line DE, a Fig. 31. perpendicular AB be drawn to that line, and also different oblique lines AE, AC, AD, &c., to different points of the same line; 1. The perpendicular AB is less than any one of the oblique lines; 2. The two oblique lines AC, AE, which meet the line DE on opposite sides of the perpendicular, and at equal distances BC, BE, from it, are equal to one another; 3. Of any two oblique lines AC, AD, or AE, AD, that which is more remote from the perpendicular is the greater. Demonstration. Produce the perpendicular AB, so that BF BA, and join FC, FD. 1. The triangle BCF is equal to the triangle BCA; for the right angle CBF = CBA, the side CB is common, and the side BF=BA; therefore the third side CF is equal to the third side AC (36). But AF <AC + CF (40), and AB half of AF is less than AC half of AC+ CF; that is, the perpendicular is less than any one of the oblique lines. 2. If BE = BC, then, as AB is common to the two triangles ABE, ABC, and the right angle ABE ABC, the triangle ABE is equal to the triangle ABC, and AE = AC. 3. In the triangle DFA, the sum of the sides AD, DF, is greater than the sum of the sides AC, CF (41); therefore AD half of AD+DF is greater than AC half of AC+ CF, and the oblique line, which is more remote from the perpendicular, is greater than that which is nearer. 53. Corollary 1. The perpendicular measures the distance of any point from a straight line. 54. Corollary II. From the same point, there cannot be drawn three equal straight lines terminating in a given straight line; for, if this could be done, there would be on the same side of the perpendicular two equal oblique lines, which is impossible. THEOREM./ 55. If, from the point C (fig. 32), the middle of the straight line AB, a perpendicular EF be drawn; 1. each point in the perpendicular EF is equally distant from the two extremities of the line AB; 2. any point without the perpendicular is at unequal distances from the same extremities A and B. = Demonstration. 1. Since AC CB, the two oblique lines. AD, DB, are drawn to points which are at the same distance from the perpendicular. They are therefore equal (52). The same reasoning will apply to the two oblique lines AE, EB, also to AF, FB, &c. Whence each point in the perpendicular EF is equally distant from the extremities of the line AB. 2. Let I be a point out of the perpendicular; join IA, IB, one of these lines must cut the perpendicular in D; join DB, then ᎠᏴ -- ᎠᏁ. But the line IB<ID + DB and ID + DB = ID + DA = IA ; therefore IBIA; that is, any point without the perpendicular is at unequal distances from the extremities of AB. |