Then any two of its angles shall be together greater 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 equals 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. I. 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 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. 1. 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. 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 A 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 DCB, 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 -I. 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. A E B CONSTRUCTION Produce BD to meet AC in E. DEMONSTRATION 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; to each of these add EC; then the sides BA, AC, are greater than BE, EC. (ax. 4.) Again, because the two sides CE, ED, of the triangle CED, are greater than CD, add DB to each of these; then the sides CE, EB, are greater than CD, DB; (ax. 4) but it has been shown that BA, AC, are greater than BE, EC, much more then are BA, AC, greater than BD, DC. Again, because the exterior angle of a triangle is greater than the interior and opposite angle; (I. 16) therefore the exterior angle BDC of the triangle CDE is greater than CED; for the same reason, the exterior angle CEB of the triangle ABE is greater than BAC; and it has been shown that the angle BDC is greater than CEB; much more then is the angle BDC greater than the angle BAC. Therefore, if from the ends, &c. Q. E. D. PROP. XXII.-PROBLEM. To make a triangle of which the sides shall be equal to three given straight lines, but of which any two whatever must be greater than the third. (References-Prop. I. 3; post. 3; ax. 1; def. 15.) Let A, B, C be three given right lines, of which any two whatever are greater than the third -- viz., A and B together greater than C; A and C together greater than B; and B and C greater than A. It is required to make a triangle having its sides equal to A, B, C, each to each. Take a straight line DE terminated at the point D, but unlimited towards E, |