b 13. 1. b Book I. ACB are greater than the angles ABC, ACB; but ACD, ACB are together equal to two right angles; therefore the angles ABC, BCA are 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. Q. E. D. a 3. 1. The greater side of every triangle is opposite to the greater angle. Let ABC be a triangle, of Because AC is greater than b 16. 1. greater than the interior and c 5. 1. a 5. 1. C D opposite angle DCB; but ADB is equal to ABD, 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 ACB. Therefore, "the greater side," &c. Q. E. D. 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. a 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 anb 18. 1. gle ABC would be less than b B A the angle ACB; but it is not: therefore the side AC is not less Book I. than AB; and it has been shown that it is not equal to AB: therefore AC is greater than AB. Wherefore," the greater angle," &c. Q. E. D. PROP. XX. THEOR. Any two sides of a triangle are together greater See N. 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. a Produce BA to the point D, and make AD equal to AC; and join DC. Because DA is equal to AC, the angle ADC is likewise C BCD is greater than the angle ADC; and because the angle BCD of the triangle DCB is greater than its angle BDC, and that the greater side is c 19. 1. 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 that AB. Therefore, any two sides," &c. Q. E. D. If, from the ends of one side of a triangle there be See N. 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 triangled are d 20. 1. greater than the third side, the two sides BA, AE of the tri Book I. angle ABE are greater than BE. e 16. 1. See N. f 20. 1. a 3. 1. B more then are BA, AC greater than BD, DC. e 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 equal to three given straight lines, but any two whatever of these must be greater than the third'. 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, and make a DF equal to A, FG to B, and GH equal to C; and from the centre F, at the distance. b 3. Post. FD, describe the circle D b DKL; and from the KF, KG; the triangle to the three straight lines, A, B, C. ६ Because the point F is the centre of the circle DKL, FD is 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 centre of the c 15. Def. 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. At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. 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. Take in CD, CE any points D, E, and join DE; and make a the triangle AFG, the sides of which shall be equal to the three A 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 equal to the angle FAG. Therefore at b 8. 1. 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. If two triangles have two sides of the one equal See N. 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. Book I. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two DE, DF, each to each, viz. AB equal to DE, and AC to DF; but the angle BAC greater than the angle EDF; the base BC is also greater than the base EF. Of the two sides DE, DF, let DE be the side which is not greater than the other, and at the point D, in the straight line a 23. 1. DE, make a the angle EDG equal to the angle BAC; and make DG equal to AC or DF, and join EG, GF. b 3. l. c 4. 1. d 5. 1. b Because AB is equal to DE, and AC to DG, the two sides BA, AC are equal to the two ED, DG, each to each, and the angle BAC is equal с to the angle EDG; A d D the angle DGF; but the angle DGF is B greater than the an gle EGF; therefore the angle DFG is greater than EGF; and much more is the angle EFG greater than the angle EGF; and because the angle EFG of the triangle EFG is greater than its angle EGF, e 19. 1. and because the greater side is opposite to the greater angle; that side EG is therefore greater than the side EF; but EG is equal to BC; and therefore also BC is greater than EF. Therefore, "if two triangles," &c. Q. E. Ď. PROP. XXV. THEOR. If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other; the angle also contained by the sides of that which has the greater base, shall be greater than the angle contained by the sides equal to them of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB equal to DE, and AC to DF; but let the base CB be greater than the base EF; the angle BAC is likewise greater than the angle EDF. |