THEOREM 7. 55. Any two sides of a triangle are together equal to more than a third side. For, in the triangle, ABC, the straight line BC is the shortest % line that can be drawn from B to C; therefore the sum of the two sides, BA, AC, is greater than BC. Theorem 5. 56. If from any point, as O, within a triangle, ABC, there be drawn two straight lines, OB, OC, one to each extremity of any side, as BC, their sum is less than the sum of the other two sides of the triangle. Produce BO till it meet AC in D; the line OC is less than the sum of the two lines OD, DC (theorem 7); and, adding to these unequals the line BO, the sum of the two lines, BO, OC, is less than the sum of the three lines BO, OD, DC {ax. 4, p. 15); that is, the sum of the two lines BO, OC, is less than the sum of the two lines BD, DC. In like manner, BD is less than the sum of the two lines BA, AD; and, adding DC to these unequals, the sum of the two straight lines, BD, DC, is less than the sum of the three straight lines BA, AD, DC; that is, the two straight lines BD, DC, are less than the two straight lines BA, AC; but the two straight lines BO, CC, have been shown to be less than the two straight lines BD, DC; and, therefore, much less is the sum of the two straight lines BO, CC, than that of the two sides BA, AC, of the triangle ABC. Theorem 9. 57. If any two sides AB, AC, of a triangle, ABC, are equal to two sides DE, DF, of another triangle DEF, each to each, and if the angle BAC, contained by the sides, AB, AC, be greater than the angle EDF, contained by the sides ED, DF, the base BC of the triangle which has the greater angle shall be greater than the base EF of the other triangle. Make the angle CAG equal to D, take AG equal to DE or AB, and join CG; and because the two triangles CAG, DEF, have an angle of the one equal to an angle of the other, and the sides which contain these angles are equal, CG shall be equal to EF Case 1.—Because GC is less than the sum of the two straight lines GI, IC; and AB less than the sum of the two straight lines AI, IB: therefore, the sum of the two straight lines GC, AB, is less than the sum of the four straight lines GI, IC, AI, IB; that is, the sum of the two straight lines GC, AB, is less than the sum of the two straight lines AG, BC; but AG is equal to AB, therefore GC is less than BC; but GC = EF, therefore EF is less than BC. Case 2.—If the point G fall on BC, it is evident that GC, or its equal EF, is less than BC. Case 3.—Lastly, if the point G fall within the triangle ABC, by theorem 8, we have the sum of the two straight lines AG, GC, less than the sum of the two straight lines mt? AB, BC; but since AB is equal to AG, we shall have GC less than BC; and, consequently, EF less than BC. Theorem 10. 58. One triangle is equal to another, when the three sides of the first are respectively equal to the three sides of the second. Let the side AB be equal to DE, AC equal to DF, and BC equal to EF; then shall the angle A be equal to the angle D, the angle B equal to the angle E, and the angle C equal to the angle F. For, if the angle 3i c E A were greater than D, then, as the two sides AB, AC, are equal to the two sides DE, DF, each to each, it would follow {theorem 9) that the side BC. would be greater than EF; and, if the angle A were less than the angle D, BC would be less than EF; therefore, the angle A can neither be greater nor less than the angle D; the angle A must therefore be equal to the angle D. In like manner, it may be proved that the angle B is equal to E, and C equal to F. 59. Corollary.—Whence it appears that, in two equal triangles the equal angles are opposite to the equal sides; for the equal angles A and D are opposite to the equal sides BC and EF. THEOREM 11. 60. The angles opposite to the equal sides of an isosceles triangle are equal. Let the side AB be equal to AC; then shall the angle C equal the x angle B. For, suppose AD to be drawn from the vertex A to the middle point D, of the base BC; then the two triangles ADB, ADC, will have the two sides AB, BD, of the one equal to the two sides :sf— , AC, CD, of the other, each to each; and AD is common to both: therefore the angle B shall be equal to the angle C. 61. Corollary 1.—Hence every equilateral triangle is also equiangular. 62. Corollary 2.—A straight line drawn from the vertex of an isosceles triangle to the middle of the base will bisect the vertical angle, and be perpendicular to the base. Theorem 12. 63. If two angles of a triangle be equal, the opposite sides shall be equal, and the triangle shall be isosceles. Let the angle ABC be equal to ACB, the side AC shall be equal to the side AB. For, if the two sides AB, AC, are not equal, let AB be greater than AC, and from BA cut off BD, equal to CA, and join CD; the ^— —^ angle DBC is, by hypothesis, equal to the angle ACB, and the two sides DB, BC, are equal to the two sides AC, CB; therefore the triangle DBC is equal to the triangle ACB, the less to the greater, which is impossible (ax. 9, p. 15); therefore AB cannot be unequal to AC, but must be equal to it. Theorem 13. 64. From a point A, without a straight line DE, only one perpendicular can be drawn to that line. For, suppose it were possible to draw AB, AC, perpendi- cular from the same point A, upon the straight line DE; produce one of them, AB, to F, so that BF may be equal to AB, and join FC; and, because AB is equal to BF, and BC is common to the two triangles ABC, BBC, and the angles ABC and FBC are equal; the angle ACB is equal to FCB Theorem 14. 65. Of all the lines that can be drawn from a given point A, to a given straight line DE, the perpendicular is the shortest; and of the other lines, that which,is nearer the perpendicular is less than that which is more remote; and those two lines, on opposite sides, and at equal distances, from the perpendicular, are equal. Produce the perpendicular, so that BF may be equal to AB, and draw the straight lines AC, AD, and AE, to meet DE in C, D, and E, and join FC, FD, &c The triangles BCF and BCA are equal (theorem, 5); for BF is equal to BA, and BC common; therefore CF is equal to CA. Now AF is less than AC+CF (theorem 7); therefore, taking the halves, AB is less than AC; that is, the perpendicular is the shortest line that can be drawn from A to DE. Next, suppose BE equal to BC; then the triangles ABE and ABC will be equal |