Book I. a 10. 1. PROP. XVI. THEOR. Ir one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles. Let ABC be a triangle, and let its side BC be produced to D, the exterior angle ACD is greater than either of the interior opposite angles CBA, BAC. Bisect AC in E, join BE and produce it to F, and make EF equal to BE; join also 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 15. 1. AEB, is equal to the angle CEF, because they are opposite vertical angles; therefore the base AB is 4. 1. 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 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 BAE: In the same manner, if the side BC be bisected, it may be 15. 1. demonstrated that the angle BCG, that is, the angle ACD, is greater than the angle ABC. Therefore, if one side, &c. Q.E.D. PROP. XVII. THEOR. ANY two angles of a triangle are together less than two right angles. Let ABC be any triangle; any two of its angles together are less than two right angles. Produce BC to D; and because ACD is the exterior angle of the triangle ABC, ACD is * 16. 1. greater than the interior and opposite angle ABC; to each of A C b these add the angle ACB; therefore the angles ACD, ACB Book I. are greater than the angles ABC, ACB; but ACD, ACB are together equal to two right angles; therefore the angles ↳ 13. 1. ABC, BCA are less than two right angles. In like manner, it may be demonstrated, that BAC, ACB, as alsó CAB, ABC, are less than two right angles. Therefore any two angles, &c. Q. E. D. PROP. XVIII THEOR. THE greater side of every triangle is opposite to the greater angle. Let ABC be a triangle, of which the side AC is greater than the side AB; the angle ABC is also greater than the angle BCA. Because AC is greater than AB, makea AD equal to AB, B and join BD; and because a 3.1. b ADB is the exterior angle of the triangle BDC, it is greaterb ↳ 16. 1. than the interior and opposite angle DCB; but ADB is equal to ABD, because the side AB is equal to the side 5.1. AD; therefore the angle ABD is likewise greater than the angle ACB. Wherefore much more is the angle ABC greater than ACB. Therefore the greater side, &c. Q. E. D. PROP, XIX. THEOR. THE greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it. Let ABC be a triangle, of which the angle ABC is greater than the angle BCA; the side AC is likewise greater than the side AB. For, if it be not greater, AC must either be equal to AB, or less than it; it is not equal, because then the angle ABC would be equal to the angle ACB; but it is not; therefore AC is not equal to AB; neither is it less; because then the B a 5.1. BOOK I. angle ABC would be less than the angle ACB; but it is not; therefore the side AC is not less than AB; and it has 48.1. been shown that it is not equal to AB; therefore AC is greater than AB. Wherefore the greater angle, &c. Q.E.D. See N. 3.1.. PROP. XX. THEOR. ANY two sides of a triangle are together greater than the third side. Let ABC be a triangle; any two sides of it together are greater than the third side, viz. the sides BA, AC greater than the side BC; and AB, BC greater than AC; and BC, CA greater than AB. Produce BA to the point D, and makea AD equal to Because DA is equal to AC, BCD is greater than the angle ACD; therefore the angle BCD is greater than the angle ADC; and because the angle BCD of the triangle DCB is - 19. 1. greater than its angle BDC, and that the greater side is opposite to the greater angle, therefore the side DB is greater than the side BC; but DB is equal to BA and AC; therefore the sides BA, AC are greater than BC. In the same manner it may be demonstrated, that the sides AB, BC, are greater than CA, and BC, CA greater than AB. Therefore any two sides, &c. Q.E.D. See N. PROP. XXI. THEOR. Ir from the ends of the side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides 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 side BC of the triangle ABC, to the point D within it; BD and DC are less than the other two sides 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 side, the two sides BA, AE, of the triangle ABE are greater than BE. To each of these EOOK I. add EC; therefore the sides BA, AC are greater than BE, EC. Again, because the two sides CE, ED of the triangle CED are greater than CD, add DB to each of these ; therefore the sides CE, EB are greater than CD, DB; but it has been shown that 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 same 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. PROP. XXII. PROB. To make a triangle of which the sides shall be See N. equal to three given straight lines, but any two whatever of these must be greater than the third a. a 20. 1. Let A, B, C be the three given straight 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 sides shall be equal to A, B, C, each to each. Take a straight line DE terminated at the point D, but unlimited towards E, HLK; and join KF, scribe another circle KG; the triangle KFG has its sides equal to the three straight lines ABC. Book I. c 15 Def Because the point F is the centre of the circle DKL, FD, is equal to FK;. but FD is equal to the straight line A; therefore 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 also 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 sides KF, FG, GK, equal to the three given straight lines A, B, C. Which was to be done. PROP. XXIII. PROB. Ar a given point in a given straight line, to make a C Α AA Take in CD, CE any points D, E, and join D4 E B/ a 22.1. DE; and make the triangle AFG, the sides of which shall be equal to the three straight lines CD, DE, EC, so 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 base DE to the base FG; the angle DCE is b8. 1. equal to the angle FAG. Therefore, at the given point A in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Which was to be done. PROP. XXIV. THEOR. See N. IF two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them, of the other; the base of that which has the greater angle shall be greater than the base of the other. |