Book I. Therefore the triangle KFC is defcribed of three ftraight lines, which are equal to the three given straight lines A, B, C. Which was to be done. With a given straight line, and at a point in it, to make a rectilineal angle equal to a given rectilineal angle. Let AB be the given straight line, and the point A the point in it; and DCE the given rectilineal angle: but it is required with the given straight line AB and at the point A in it to make a rectilineal angle equal to the given rectilineal angle DCE. Take in each of the lines CD, CE any points whatever D and E, and join DE, and (by prop. 22.) describe the triangle AGF of three straight lines, which are equal to the three, CD, DE, CE; fo that CD be equal to AF; CE to AG, and alfo DE to FG. D C A F E G /B Since therefore the two DC, CE are equal to the two FA, AG, each to each, and the base DE equal to the base FG; therefore (by prop. 8.) the angle DCE is equal to the angle FAG. Wherefore with a given straight line AB, and at a point in it the point A, a rectilineal angle FAG is made equal to the given rectilineal angle DCE. Which was to be done. PROP. XXIV. If two triangles have the two fides equal to the two fides, each to each, but have the angle greater than the angle, the angle contained by the equal ftraight lines: alfo they will have the bafe greater than the base. Let there be two triangles the triangles ABC, DEF having the two fides AB, AC equal to the two fides DE, DF, each to each; AB to DE, and AC to DF; but let an angle the angle contained by BAC be greater than the angle contained by EDF; I fay that the base BC is greater than the base EF. For Since therefore AB is equal to DE and AB to DG; certainly the the two BA, AC are equal to the two ED, DG, each to each; and the angle BAC is equal (by conft.) to EDG therefore the base BC is equal to the bafe EG. Again because DG is equal to DF, the angle DFG is equal (by prop. 5.) to the angle DGF; therefore the angle DFG is greater than the angle EGF; therefore the angle EFG is greater by much than the angle EGF; and because there is a triangle, the triangle EFG, having the angle EFG greater than the angle EGF; but (by prop. 19.) the greater fide is extended under the greater angle: therefore the fide EG is greater than EF: and EG is equal to BC (by part. 1. of this prop.); wherefore alfo BC is greater than EF. Wherefore if two triangles have the two fides equal to the two fides, each to each, and have the angle greater than the angle, the angle contained by the equal straight lines; they will also have the bafe greater than the bafe. Which was to be demonstrated. If two triangles have the two fides equal to the two fides each to each, and have the base greater than the base; they will also have the angle greater than the angle, the angle contained by the equal ftraight lines. Let there be two triangles the triangles ABC, DEF having the two fides BA, AC equal to the two fides DE, DF, each to each, AB to DE and AC to DF; but let the bafe BC be greater than the bafe EF: I fay alfo that an angle, the angle BAC is greater than an angle, the angle EDF. VOL. I. D For base EF: but it is not (by fupp.); therefore the angle BAC is not lefs than EDF: but it has been demonftrated that neither is it equal wherefore the angle BAC is greater than EDF, Wherefore if two triangles have the two fides equal to the two fides each to each, and have the base greater than the base; they will also have the angle greater than the angle, the angle contained by the equal ftraight lines. Which was to be demonstrated. If two triangles have the two angles equal to the two angles, each to each, and one fide equal to one fide; either the fide which is at the equal angles, or that which is extended under one of the equal angles: they will also have the two remaining fides equal to the two remaining fides each to each, and the remaining angle equal to the remaining angle. Let there be two triangles the triangles ABC, DEF having the two angles ABC, BCA equal to the two angles DEF, EFD each to each; the angle ABC equal to the angle DEF; and the angle BCA equal to EFD: and let them have also one fide equal to one fide: First, the fide at the equal angles, the ftraight line BC equal to EF: I fay also that they will have the two remaining fides equal to the two remaining fides, each to each; AB equal to DE and AC to DF; and the remaining angle equal to the remaining angle, the angle BAC equal to EDF. For if AB is unequal to DE, one of them will be greater than the other let AB be the greater and make GB equal (by prop. 3.) to DE, and let CG be joined. Where Wherefore because BG is equal to DE (by conft.) and BC to EF Book I. (by fupp.) certainly the two BG, BC are equal to the two DE, EF each to each; and the angle GBC is equal (by fupp.) to DEF: therefore (by prop. 4.) the base GC is equal to the base DF; and the triangle GCB will be equal to the triangle DEF; and the remaining angles will be equal to the remaining angles, each, to each, under which the equal fides are extended: therefore the angle GCB is equal to the angle DFE; but the angle DFE is fuppofed equal to BCA; therefore BCG is equal to BCA the lefs to the greater, which is impoffible; therefore AB is not unequal to DE; therefore equal but BC is also equal to EF; therefore the two AB, BC are equal to the two DE, EF, each to each, and the angle ABC is equal to the angle DEF; wherefore the base AC is equal to the base DF and the remaining angle BAC is equal to the remaining angle EDF. But again, let the fides extended under the equal angles be equal, as AB equal to Α G D DE: I fay again that will be equal to the re- B HC E F DF and BC to EF; and befides the remaining angle BAC is For if BC be unequal to EF, one of them is greater than the And fince BH is equal to EF (by conft.) and AB to DE (by supp.); certainly the two AB, BH is equal to the two DE, EF, each to each, and they contain equal angles; therefore (by prop. 4.) the base AH is equal to the base DF, and the triangle ABH is equal to the triangle DEF, and the remaining angles will be equal to the remaining angles, each to each, under which the equal fides are extended; wherefore the angle BHA is equal to EFD; but the angle EFD is equal (by fupp.) to the angle BCA; and (by com. not. 1.) BHA is therefore equal to BCA; the outward angle of the triangle AHC viz. BHA equal to the inward and opposite. D 2 BCA Book I. BCA, which (by prop. 16.) is impoffible: wherefore BC is not unequal to EF, therefore it is equal: but AB is also equal to DE; therefore the two AB, BC are equal to the two DE, EF and they contain equal angles; therefore (by prop. 4.) the base AC is equal to the bafe DF, and the triangle ABC is equal to the triangle DEF and the remaining angle BAC is equal to the remaining angle EDF. If therefore two triangles have the two angles equal to the two angles, each to each, and have one fide equal to one fide; either the fide which is at the equal angles, or that which is extended under one of the equal angles: they will also have the two remaining fides equal to the two remaining fides, each to each, and the remaining angle equal to the remaining angle. Which was to be demonftrated. If a straight line falling upon two straight lines makes the alternate angles equal to one another, the straight lines will be parallel to one another. For let the straight line EF falling upon the two straight lines AB, CD make the alternate angles AEF, EFD equal to one another. I fay that the ftraight line AB is parallel to the straight line. CD. A E B For if not, AB and CD being produced will meet, either towards the the parts B, D or towards A, C: let them be produced and let them meet towards the parts B, D in the point G. Certainly (by prop. 16.) the. outward angle AEF of the triangle EFG is greater than the inward and oppofite angle EFG; but it is alfo (by supp.) equal, which is impoffible: therefore AB, CD being produced, will not meet towards the parts B, D; certainly in the fame manner it will be C F D A E B G D F shewn, that neither will they meet towards A, C; but thofe ftraight |