A the straight line AD, making the angle BAD equal to DAC. Then the two triangles BAD, CAD have the two sides A B, AD and the included angle in the one equal to the two sides A C, AD and the included angle in the other, each to each; hence the two triangles are equal, and the angle B is equal to the angle C (Prop. V.). B 57. Cor. 1. The line bisecting the vertical angle of an isosceles triangle bisects the base at right angles. 58. Cor. 2. Conversely, the line bisecting the base of an isosceles triangle at right angles, bisects also the vertical angle. 59. Cor. 3. Every equilateral triangle is also equiangular. PROPOSITION VIII.-THEOREM. Α 60. If two angles of a triangle are equal, the opposite sides are also equal, and the triangle is isosceles. Let A B C be a triangle having the angle B equal to the angle C; then will the side A B be equal to the side A C. B D C For, if the two sides are not equal, one of them must be greater than the other. Let A B be the greater; then take DB equal to AC the less, and draw CD. Now, in the two triangles DBC, ABC, we have D B equal to AC by construction, the side BC common, and the angle B equal to the angle ACB by hypothesis; therefore, since two sides and the included angle in the one are equal to two sides and the included angle in the other, each to each, the triangle DBC is equal to the triangle ABC (Prop. V.), a part to the whole, which is impossible (Art. 34, Ax. 8). Hence the sides AB and AC cannot be unequal; therefore the triangle ABC is isosceles. 61. Cor. Therefore every equiangular triangle is equilateral. 62. Any side of a triangle is less than the sum of the other two. In the triangle A B C, any one side, as AB, is less than the sum of the other two sides, A C and CB. A C B For the straight line AB is the shortest line that can be drawn from the point A to the point B (Art. 34, Ax. 10); hence the side AB is less than the sum of the sides AC and C B. In like manner it may be proved that the side AC is less than the sum of A B and B C, and the side BC less than the sum of B A and A C. 63. Cor. Since the side AB is less than the sum of A C and C B, if we take away from each of these two unequals the side CB, we shall have the difference between AB and CB less than AC; that is, the difference between any two sides of a triangle is less than the other side. PROPOSITION X.-THEOREM. 64. The greater side of any triangle is opposite the greater angle. In the triangle CAB, let the angle Α C be greater than B; then will the side A C, opposite to B. Draw the straight line CD, making the angle B C D equal to B. Then, in C the triangle BDC, we shall have the D B side BD equal to DC (Prop. VIII.). But the side A C is less than the sum of AD and DC (Prop. IX.), and the sum of A D and D C is equal to the sum of AD and D B, which is equal to A B; therefore the side AB is greater than A C. 65. Cor. 1. Therefore the shorter side is opposite to the less angle. 66. Cor. 2. In the right-angled triangle the hypothenuse is the longest side. PROPOSITION XI.-THEOREM. 67. The greater angle of any triangle is opposite the greater side. In the triangle CA B, suppose the A side A B to be greater than AC; then will the angle C, opposite to AB, be greater than the angle B, opposite to A C. C B For, if the angle C is not greater than B, it must either be equal to it or less. If the angle C were equal to B, then would the side A B be equal to the side A C (Prop. VIII.), which is contrary to the hypothesis; and if the angle C were less than B, then would the side AB be less than AC (Prop. X. Cor. 1), which is also contrary to the hypothesis. Hence, the angle C must be greater than B. 68. Cor. It follows, therefore, that the less angle is opposite to the shorter side. 69. If, from any point within a triangle, two straight lines are drawn to the extremities of either side, their sum will be less than that of the other two sides of the triangle. Let the two straight lines BO, CO be drawn from the point O, within the triangle ABC, to the extremities of the side BC; then will the sum of the two lines BO and OC be less than the sum of the sides BA and A C. B C Let the straight line BO be produced till it meets the side AC in the point D; and because one side of a triangle is less than the sum of the other two sides (Prop. IX.), the side OC in the triangle CDO is less than the sum of OD and D C. To each of these inequalities add B O, and we have the sum of BO and O C less than the sum of BO, OD, and DC (Art. 34, Ax. 4); or the sum of BO and O C less than the sum of BD and D C. Again, because the side BD is less than the sum of B A and A D, by adding DC to each, we have the sum of BD and DC less than the sum of BA and A C. But it has been just shown that the sum of BO and OC is less than the sum of BD and DC; much more, then, is the sum of BO and OC less than BA and A C. PROPOSITION XIII.-THEOREM. 70. From a point without a straight line, only one perpendicular can be drawn to that line. Let A be the point, and DE the given straight line; then from the point A only one perpendicular can be drawn to DE. Let it be supposed that we can draw two perpendiculars, AB and AC. Produce one of them, as A B, till BF is equal to A B, and join F C. D A E C B F Then, in the triangles A B C and CBF, the angles CBA and CBF are both right angles (Prop. I. Cor. 1), the side C B is common to both, and the side BF is equal to A D E CB the side A B; hence the two triangles are equal, and the angle BCF is equal to the angle BCA (Prop. V.) But the angle BCA is, by hypothesis, a right angle; therefore BCF must also be a right angle; and if the two adjacent angles, BCA and BCF, are together equal to two right angles, the two lines AC and CF must form one and the same straight line (Prop. II.). Whence it follows, that between the same two points, A and F, two straight lines can be drawn, which is impossible (Art. 34, Ax. 11) ; hence no more than one perpendicular can be drawn from the same point to the same straight line. F E D 71. Cor. At the same point C, in the line A B, it is likewise impossible to erect more than one perpendicular to that line. For, if CD and CE were each perpendicular to A B, the angles BCD, BCE would be right angles; hence the angle BCD would be equal to the angle BCE, a part to the whole, which is impossible. A PROPOSITION XIV.-THEOREM. B C 72. If, from a point without a straight line, a perpendicular be let fall on that line, and oblique lines be drawn to different points in the same line; 1st. The perpendicular will be shorter than any oblique line. 2d. Any two oblique lines, which meet the given line at equal distances from the perpendicular, will be equal. 3d. Of any two oblique lines, that which meets the given line at the greater distance from the perpendicular will be the longer. |