Book I. a. 15. I. IF PROP. XVI. THEOR. F one fide of a triangle be produced, the exterior angle is greater than either of the interior oppo'fite angles. Let ABC be a triangle, and let its fide 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 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 to the angle CEF, because they are opposite verti cal angles. therefore the bafe C. 4. I.. AB is equal to the base CF, and the triangle AEB to the 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 ACD is greater than BAE. in the fame manner, if the fide BC be bifected, it may be demonftrated that the angle d. 15. t. BCG, that is, the angle ACD, is greater than the angle ABC. therefore if one fide, &c. Q. E. D. 2. 16. 1. A PROP. XVII. THEOR. NY two angles of a triangle are together less than two right angles. Let ABC be any triangle; any two of its angles together are lefs than two right angles. Produce BC to D; and because ACD is the exterior angle of the triangle ABC, ACD is greater than the interior and oppofite angle ABC; to each of a A thefe add the angle ACB, therefore the angles ACD, ACB are Book I. greater than the angles ABC, ACB. but ACD, ACB are together b equal to two right angles; therefore the angles ABC, BCA are b. 13. 1. less than two right angles. in like manner it may be demonstrated that BAC, ACB, as also CAB, ABC are less than two right angles. therefore any two angles, &c. PROP. XVIII. Q. E. D. THEOR. HE greater fide of every triangle is oppofite to the Let ABC be a triangle of which the fide AC is greater than the fide AB; the angle ABC is also greater than the angle BCA. Because AC is greater than AB; make a AD equal to AB, and join BD. and becaufe ADB B is the exterior angle of the tri b A a. 3. I. C. 5. I. angle BDC, it is greater than the interior and oppofite angle b. 16. 1. DCB. but ADB is equal to ABD, because the fide 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. Q. E. D. THE HE greater angle of every triangle is fubtended by 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 muft either be equal to AB, or lefs than it. it is not equal, because then the angle ABC would be equal to the angle ACB; but it a is not; therefore AC is not equal to AB. neither is it lefs; because then the angle ABC would be lefs A a. 5. I. Book Ib than the angle ACB; but it is not; therefore the fide AC is ---— not less than AB. and it has been fhewn that it is not equal to b. 18. 1. AB. therefore AC is greater than AB. wherefore the greater angle, &c. Q. E. D. See N. a. 3. I. b. 5. I. C. 19. I. See N. A PROP. XX. THEOR. NY two fides of a triangle ate 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. Because DA is equal to AC, the angle ADC is likewife equal b to ACD. but the angle BCD is greater than the angle ACD; therefore the angle BCD is great D A. er than the angle ADC. and because the angle BCD of the triangle DCB is greater than its angle BDC, and that the greater c 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 demonftrated that the fides AB, BC are greater than CA; and BC, CA greater than AB. therefore any two fides, &c. Q. E. D. IF F from the ends of the fide of a triangle there be drawn two ftraight lines to a point within the triangle, these shall be less than the other two fides of the triangle, but shall contain a greater angle. Let the two straight 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. Produce BD to E; and because two sides 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 Book I. BA, AC are greater than BE, EC.'again, because the two fides A CE, ED of the triangle CED are greater than CD, add DB to E each of thefe; therefore the fides CE, EB are greater than CD, BA, AC are greater than BE, B 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, the exterior angle BDC of the triangle CDE is greater than CED. for the fame reason, the exterior angle CEB of the triangle ABE is greater than BAC. and it has been demonstrated that the angle 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. O make a triangle of which the fides fhall be equal See N. to three given straight lines; but any two whatever of these must be greater than the third a. 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 straight line DE terminated at the point D, but unli a. 20. X. Because the point F is the center of the circle DKL, FD is c. 15. Def. C c Book I. equal to FK; but FD is equal to the straight line A; therefore FK is equal to A. again, because G is the center of the circle LKH, GH is equal to GK; but GH is equal to C, therefore also GK is equal to C. and FG is equal to B; therefore the three flraight 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 straight lines A, B, C. Which was to be done, 8. 22. I. 1. 8. I. ee N. A Ta given point in a given ftraight line to make a rectilineal angle equal to a given rectilineal angle. A Let AB be the given straight line, and A the given point in it, and DCE the given rectilineal angle; it is required to make an angle at the given point A in the given straight line AB that shall be equal to the given rectilineal angle DCE. three ftraight lines CD, DE, EC, fo that CD be equal to AF, CE to AG, and DE to FG. and because DC, CE are equal to FA, AG, each to each, and the bafe DE to the bafe FG; the angle DCE is equal to the angle FAG. therefore at the given point A in the given ftraight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Which was to be done. IF F two triangles have two fides of the one equal to two fides of the other, each to each, but the angle contained by the two fides of one of them greater than the angle contained by the two fides equal to them, of the other; the bafe of that which has the greater angle fhall be greater than the base of the other. Let ABC, DEF be two triangles which have the two fides AB, |