Book I. a 10. I. PRO P. XVI. THE OR. F one fide of a triangle be produced, the exterior I angle is greater than either of the interior oppo fite angles. Let ABC be a triangle, and let its fides BC be produced to D, the exterior angle ACD is greater than either of the interior oppofite angles CBA, BAC. Bifecta AC in E, join BE and produce it to F, and make EF equal to BE; join alfo FC, and produce AC to G. Because AE is equal to EC,and BE to EF, AE, EB are equal to CE, EF, each to each; and the angle AEB is b 15. 1. equal b to the angle CEF, because they are oppofite vertical angles; therefore the base C 4.1.. AB is equal to the base CF, and the triangle AEB to the B A. E F triangle CEF, and the remaining angles to the remaining angles, each to each, to which the equal fides are oppofite wherefore the angle BAE is equal to the angle ECF; but the angle ECD is greater than the angle ECF; therefore the angle ECD, that is ACD, is greater than BAE: In the fame manner, if the fide BC be bifected, it may be demonftrated d15 1. that the angle BCG, that is d, the angle ACD, is greater than the angle ABC. Therefore, if one fide, &c. Q. E. D. A NY two angles of a triangle are together lef than two right angles. Le Let ABC be any triangle; any two of its angles together are lefs than two right angles. Produce BC to D; and becaufe ACD is the exterior angle of the triangle ABC, ACD is greater a than the interior and oppofite angle ABC; to each of these add the angle ACB; therefore the angles ACD, ACB are greater than the angles ABC, ACB; but ACD, ACB are together equal b to two right b 13. 1. angles; therefore the angles ABC, BCA are lefs than two right angles. In like manner, it may be demonftrated, that BAC, ACB, as alfo CAB, ABC, are lefs than two right angles. Therefore, any two angles, &c. QE. D. PRO. P. XVIII. THE O R. HE greater fide of every triangle has the great-. TH Let ABC be a triangle, of Because AC is greater than C AAT a 3. I. and join BD; and because ADB is the exterior angle of the riangle BDC, it is greater b than the interior and oppofite b 16. 1. angle DCB; but ADB is equal to ABD, because the fide c 5. 1. AB is equal to the fide AD; therefore the angle ABD is likewife greater than the angle ACB; wherefore much more Is the angle ABC greater than ACB. Therefore the greater fide, &c. QE. D. Book I. 2 5.1. T HE greater angle of every triangle is fubtended by the greater fide, or has the greater fide oppofite to it. Let ABC be a triangle, of which the angle ABC is greater than the angle BCA; the fide AC is likewife greater than the fide AB. For, if it be not greater, AC b18. 1. ABC would be lefs b than the angle ACB; but it is not; therefore the fide AC is not lefs than AB; and it has been fhewn that it is not equal to AB; therefore AC is greater than AB. Wherefore the greater angle, &c. Q. E. D. a 3.1. b 5. I. PRO P. XX. THE O R. ANY two fides of a triangle are together greater than the third fide. Let ABC be a triangle; any two fides of it together are greater than the third fide, viz. the fides BA, AC greater than the fide BC; and AB, BC greater than AC; and BC, CA greater than AB. Produce BA to the point D, and make a AD equal to AC; and join DC. Becaufe DA is equal to AC, the angle ADC is likewife equal to ACD; but the angle BCD is greater than the angle ACD there fore the angle BCD is greater than the angle ADC; and be caufe c 19. 1. caufe the angle BCD of the triangle. DCB is greater than its Book I angle BDC, and that the greater & fide is oppofite to the greater angle; therefore the fide DB is greater than the fide BC; but DB is equal to BA and AC; therefore the fides BA, AC are greater than BC. In the fame manner it may be demonstrated, that the fides AB, BC are greater than CA, and BC, CA greater than AB. Therefore any two fides, &c. Q. E. D. IF PROP. XXI. THE OR. F from the ends of the fide of a triangle, there be See N drawn two ftraight lines to a point within the triangle, these two lines fhall be less than the other two fides of the triangle, but fhall contain a greater angle. Let the two ftraight lines BD, CD be drawn from B, C, the ends of the fide BC of the triangle ABC, to the point D within it; BD and DC are lefs than the other two fides BA, AC of the triangle, but contain an angle BDC greater than the angle BAC. E D Produce BD to E; and because two fides of a triangle are greater than the third fide, the two fides BA, AE of the triangle ABE are greater than BE. To each of these add EC; therefore the fides BA, AC are greater than BE, EC: Again, because the two fides CE, ED of the triangle CED are greater than CD, add DB to each of thefe; therefore the fides CE, EB are greater than CD, DB; but it has been shewn that BA, AC are greater than BE, EC; much more then are BA, AC greater than BD, DC. B C Again, because the exterior angle of a triangle is greater than the interior and oppofite angle, the exterior angle BDC of the triangle CDE is greater than CED; for the fame reafon, the exterior angle CEB of the triangle ABE is than BAC; and it has been demonftrated that the angle G 4 greater BDC Book I. BDC is greater than the angle CEB; much more then is the angle BDC greater than the angle BAC. Therefore, if from the ends of, &c. Q. E. D. To make a triangle of which the fides fhall be equal to three given ftraight lines; but any two whatever of thefe lines must be greater than the á 25. 1. third a. a 3. I. 13 Let A, B, C be the three given ftraight lines, of which any two whatever are greater than the third, viz. A and B greater than C; A and C greater than B; and B and C than A. It is required to make a triangle of which the fides fhall be equal to A, B, C, each to each. Take a ftraight line DE terminated at the point D, but unlimited towards E, and make a DF equal to A, FG to B, and GH equal to C; and from the centre F, at the distance FD, de- D Poft. fcribeb the circle DKL; and from the centre G, Because the point F is the centre of the circle DKL, FD is e 11. Def. equal to FK, but FD is equal to the ftraight line A; there fore FK is equal to A: Again, because G is the centre of the circle LKH, GH is equal to GK; but GH is equal to C; therefore alfo GK is equal to C; and FG is equal to B; therefore the three ftraight lines KF, FG, GK, are equal to the three A, B, C: And therefore the triangle KFG has its three fides KF, FG, GK equal to the three given ftraight lines, A, B, C. Which was to be done. PROP. |