equal to two right angles; therefore the angles CBA, ABE are Book I. equal to the angles CBA, ABD: Take away the common angle ABC, and the remaining angle ABE is equal b to the b 3. Ax, remaining angle ABD, the lefs to the greater, which is impoffible; therefore BE is not in the fame ftraight line with BC. And, in like manner, it may be demonftrated, that no other can be in the fame ftraight line with it but BD, which therefore is in the fame ftraight line with CB. Wherefore, if at a point, &c. Q. E. D. I' F two ftraight lines cut one another, the vertical, or oppofite, angles fhall be equal. Let the two ftraight lines AB, CD cut one another in the point E; the angle AEC fhall be equal to the angle DEB, and CEB to AED. C a 13. I. A E B D Because the ftraight line AE makes with CD the angles CEA, AED, these angles are together equal to two right angles. Again, because the straight line DE makes with AB the angles AED, DEB, thefe alfo are together equal a to two right angles; and CEA, AED have been demonftrated to be equal to two right angles; wherefore the angles CEA, AED are equal to the angles AED, DEB. Take away the common angle AED, and the remaining angle CEA is equal b to the remaining angle DEB. In the fame manner, it can be demonftrated, that the angles CEB, AED are equal. Therefore, if two ftraight lines, &c. Q. E. D. In the b3. Az, COR. 1. From this it is manifeft, that, if two ftraight lines cut one another, the angles they make at the point where they cut, are together equal to four right angles. COR. 2. And confequently, that all the angles made by any number of lines meeting in one point, are together equal to four right angles. COR. 3. Likewife, if two ftraight lines CE, ED, on oppofite fides of a straight line AB, meet at a point E in it, and make the vertical angles AEC, BED equal to one another, they are in the fame ftraight line. For the angles DEB, BEC are equal to AEC, CEB, that is, to two right angles d; therefore CE is in the fame ftraight line with ED. C 2. Ax. d 13. 1. e 14. f. PROP. Book I. a 10. I. I PROP. XVI. THEOR. F one fide of a triangle be produced, the exterior angle is greater than either of the interior oppofite 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, and Because AE is equal to EC, and b A. E F C D € 4.1. angles; therefore the base AB is c A PROP. XVII. THEOR. NY two angles of a triangle are together lefs than two right angles. Let ABC be any triangle; any two of its angles together are lefs than two right angles. B A C D Produce BC to D; and because ACD is the exterior angle of the a 16. 1. 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 b 13. 1. ABC, ACB; but ACD, ACB are together equal b to two right angles; therefore the angles ABC, BCA are less than two right angles. In like manner, it may be demonstrated, that BAC, : BAC, ACB, as alfo CAB, ABC are lefs than two right angles. Book I Therefore any two angles, &c. Q. E. D. PROP. XVIII. THEOR. HE greater fide of every triangle is oppofite to THE the greater angle. Let ABC be a triangle, of which the fide AC is greater than the fide AB; the angle ABC is alfo greater than the angle BCA B A D a 3. 1. Because AC is greater than AB, make a AD equal to AB, and join BD; and because ADB is the exterior angle of the triangle BDC, it is greater than the interior and oppofite angle DCB; but ADB b 15. r. is equal to AED, 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 PROP. XIX. THEOR. THE greater angle of every triangle is fubtended by the greater fide, or has the greater fide oppo fite 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 muft either be equal to AB, or lefs than it ; it is not equal, because then the angle ABC would be equal a to the angle ACB; but it is not; therefore AC is not equal to AB: neither is it lefs; because then the angle ABC would be C 5. I. a 5. 1. less b than the angle ACB; but it is not; therefore the fide AC b 18. 1. is not less 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 NY two fides of a triangle are together greater BOOK I. a 3. I. 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 D, and make a AD equal to AC; and join DC. C Becaufe DA is equal to AC, the b 5. 1. angle ADC is likewise equal b to ACD; but the angle BCD is greater than ACD; therefore BCD is also greater than ADC; and because the angle BCD of the triangle DCB is greater than its angle B 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. C 19. I. IF PROP. XXI, THEOR. [F, from the ends of the fide of a triangle, there be drawn two ftraight lines to a point within the triangle, these fhall be less than the other two fides of the triangle, but fhall 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 less 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 fides of a triangle are a 20. 1. greater than the third fide a, 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 b. Again, b 2. Ax. because the two fides CE, ED of the triangle CED are greater than B A E D are BA, AC greater than BD, DC. Again, because the exterior angle of a triangle is greater than c 16. 1. the interior and oppofite angle, the exterior angle BDC of the triangle CDE is greater than CED; for the fame reason, the exterior exterior angle CEB of the triangle ABE is greater than BAC; Book. I. and it has been demonftrated, 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. T PROP. XXII. PROB. O make a triangle of which the fides fhall be See N. equal to three given ftraight 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 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; therefore any two of the ftraight lines DF, FG, GH are greater than the third: from the centre F, at the diftance FD, describe b the circle DKL; it fhall cut FH, for FD is lefs than FH: alfo from the centre G, at the distance GH, describe another circle HKL; and it shall cut GD, for GH is lefs than GD: and the two circles cut one another, because DF, GH are together greater than FG; let them cut one another in K, and join KF, KG; the triangle KFG has its fides equal to the three straight lines A, B, C. C a 20. I. a 3. t. b 3. Poft. Because the point F is the centre of the circle DKL, FD is equal to FK; but FD is equal to the ftraight line A; there- c 15. Def. 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 straight 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. A angle. PROP. XXIII. PROB. Ta given point in a given straight line, to make a |