Let ABC be a triangle, and let the side BC be produced to D. Then the exterior angle ACD shall be greater than either of the angles ABC or CAB. A B G CONSTRUCTION Bisect AC in E, (1. 10) and join BE, produce BE to F, and make EF equal to BE; (1.3) and join FC. DEMONSTRATION Because AE is equal to EC, and BE to EF; the two sides AE, EB, are equal to the two CE, EF, each to each; and the angle AEB is equal to the angle CEF, because they are opposite, vertical angles; (1. 15) therefore the base AB is equal to the base CF; (1. 4) and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite; wherefore the angle BAE is equal to the angle ECF; but the angle ECD is greater than the angle ECF; therefore the angle ACD is greater than the angle BAE. In the same manner, if the side BC be bisected, and AC be produced to G, it may be demonstrated that the angle BCG, that is, the angle ACD, (1. 15) is greater than the angle АВС. Therefore, if one side, &c. Q. E. D. PROP. XVII.—THEOREM. Any two angles of a triangle are together less than two right angles. (References-Prop. I. 13, 16; ax. 4.) Let ABC be any triangle. Then any two of its angles shall be together less than two right angles, B CONSTRUCTION Produce the side BC to D. DEMONSTRATION Because ACD is the exterior angle of the triangle ABC, therefore the angle ACD is greater than the interior and opposite angle ABC; (1. 16) to each of these unequals add the angle ACB; then the angles ACD, ACB, are greater than the angles ABC, ACB; but the angles ACD, ACB, are together equal to two right angles; (1. 13) therefore the angles ABC, BCA, are less than two right angles. In like manner, it may be proved, that the angles BAC, ACB, are less than two right angles, and likewise the angles CAB, ABC, Therefore any two angles, &c. Q. E. D. PROP. XVIII.—THEOREM. The greater side of every triangle is opposite to the greater angle; i.e., in any triangle, if one side be greater than another, then the angle which is opposite the greater side is greater than the angle which is opposite the less. (References - Prop. 1. 3, 5, 16.). Let ABC be any triangle of which the side AC is greater than the side AB. Then the angle ABC shall be greater than the angle ВСА. B D CONSTRUCTION Because the side AC is greater than the side AB, make AD equal to AB, (1. 3.) and join BD. DEMONSTRATION Because the angle ADB is the exterior angle of the triangle BDC, therefore the angle ADB is greater than the interior and opposite angle DCB; (1. 16) but the angle ADB is equal to the angle ABD, (1. 5) because the side AB is equal to the side AD, therefore the angle ABD is likewise greater than the angle ACB; wherefore much more is the angle ABC greater than the angle ACB. Therefore the greater side, &c. Q. E. D. PROP. XIX.—THEOREM. The greater angle of every triangle is subtended by the greater side, i.e., in any triangle, if one angle be greater than another, then the side which is opposite the greater angle is greater than the side which is opposite to the less. (References-Prop. I. 5, 18.) Let ABC be any triangle, of which the angle ABC is greater than the angle BCA. Then the side AC shall be greater than the side AB. B C For, if AC be not greater than AB, AC must either be equal to AB, or less than it. But AC is not equal to AB, because then the angle ABC would be equal to the angle ACB; (1. 5) but it is not, (hyp.) therefore AC is not equal to AB. Neither is AC less than AB; because then the angle ABC would be less than the angle ACB; (1. 18) but it is not; (hyp.) therefore AC is not less than AB; and it has been shown that it is not equal to AB; therefore AC is greater than AB. Wherefore, in any triangle, &c. Q. E. D. PROP. XX.- THEOREM. Any two sides of a triangle are together greater than the third side. (References-Prop. 1. 3, 5, 19; ax. 9.) Let ABC be a triangle. Then any two sides of it together are greater than the third side, viz., the sides BA, AC, greater than the side BC; AB, BC, greater than AC; and BC, CA, greater than AB. B CONSTRUCTION Produce BA to the point D, and make AD equal to AC; (1. 3) and join DC. DEMONSTRATION Because DA is equal to AC, therefore the angle ADC is equal to the angle ACD; (1. 5) but the angle BCD is greater than the angle ACD; therefore the angle BCD is greater than the angle ADC; and because in the triangle DCB, the angle BCD is greater than the angle BDC, and that the side which is opposite the greater angle is greater than that which is opposite the less; (I. 19) therefore the side DB is greater than the side BC; but DB is equal to BA and AC; (constr.) therefore the sides BA, AC, are greater than BC. In the same manner, it may be proved, that the sides AB, BC, are greater than CA; and BC, CA, are greater than AB. Therefore, any two sides, &c. Q. E. D. PROP. XXI.- THEOREM. If from the ends of a side of a triangle there be drawn two straight lines to a point within the triangle; then these shall be less than the other two sides of the triangle, but shall contain a greater angle. (References -1. 16, 20; ax. 4.) Let the two straight lines BD, CD, be drawn from B, C, the ends of the side BC of the triangle ABC, to the point D within it. Then BD and DC shall be less than the two sides BA, AC, of the triangle, but shall contain an angle BDC greater than the angle BAC. Because two sides of a triangle are greater than the third side, (1. 20) therefore the two sides BA, AE, of the triangle ABE, are greater than BE; |