PROPOSITION XXIII. PROBLEM. With a given Right Line, and at a given Point in it, to make a Right-lin❜d Angle equal to a Right-lin'd Angle given. LE ET the given Right Line be AB, and the Point given therein A, and the given Right-lin❜d Angle DCE. It is required to make a Right-lin'd Angle at the given Point A, with the given Right Line A B, equal to the given Right-lin'd Angle DCE. Affume the Points D and E at Pleasure in the Lines CD, CE, and draw DE; then, of three Right Lines *22 of this. equal to CD, DE, EC, make a Triangle AFG, fo that AF be equal to CD, AG to CE, and FG to DE. Then because the two Sides DC, CE, are equal to two Sides FA, AG, each to each, and the Base DE equal to the Bafe FG; the Angle DCE fhall be + 8 of this. † equal to the Angle FAG. Therefore the Rightlin'à Angle FAG is made, at the given Point A, in the given Line AB, equal to the given Right-lin❜d Angle DCE; which was to be done. PROPOSITION XXIV. THEOREM. If two Triangles have two Sides of the one, equal to two Sides of the other, each to each, and the Angle of the one, contained under the equal Right Lines, greater than the correspondent Angle of the other; then the Bafe of the one will be greater than the Bafe of the other. LE ET there be two Triangles ABC, DEF, having two Sides AB, AC, equal to the two Sides DE, DF, each to each, viz. the Side A B equal to the Side DE, and the Side AC equal to DF; and let the Angle BAC be greater than the Angle EDF. I fay, the Bafe BC is greater than the Bafe EF. For * For because the Angle B A C is greater than the Angle EDF, make an Angle EDG at the Point D *23 of this. in the Right Line DE, equal to the Angle BAC, and make DG equal to either AC or DF, and join EF, + 3 of this. FG. 5 of this, Now because AB is equal to DE, and AC to DG, the two Sides BA, AC, are each equal to the two Sides ED, DG, and the Angle BAC equal to the Angle EDG: Therefore the Bafe BC is equal to || 4 of this? the Bafe EG. Again, because DG is equal to DF, the Angle DFG is equal to the Angle DGF; and fo the Angle DFG is greater than the Angle EGF: And confequently the Angle EFG is much greater than the Angle EGF. And becaufe EFG is a Triangle, having the Angle EFG greater than the Angle EGF; and the greatest Side fubtends the greatest 19 of this. Angle, the Side EG fhall be greater than the Side EF. But the Side EG is equal to the Side BC. Whence BC is likewife greater than EF. Therefore if two Triangles have two Sides of the one, equal to two Sides of the other, each to each, and the Angle of the one, contain'd under the equal Right Lines, greater than the correfpondent Angle of the other; then the Bafe of the one will be greater than the Bafe of the other; which was to be demonftrated. PROPOSITION XXV, THEOREM. If two Triangles have two Sides of the one equal to two Sides of the other, each to each, and the Bafe of the one greater than the Bafe of the other; they shall also have the Angles, contain'd under the equal Sides, the one greater than the other. LE ET there be two Triangles ABC, DEF, having two Sides A B, A C, each equal to two Sides DE, DF, viz. the Side AB equal to the Side DE, and the Side AC to the Side DF; but the Base BC greater than the Bafe EF. I fay, the Angle BAC is alfo greater than the Angle EDF. For if it be not greater, it will be either equal or lefs. But the Angle BAC is not equal to the Angle C 4 EDF; * 4 of this. EDF; for if it was, the Bafe BC would be * equal to the Bafe EF; but it is not: Therefore the Angle BAC is not equal to the Angle EDF, neither will it † 24 of this. be leffer; for if it fhould, the Bafe BC would be + lefs than the Bafe EF; but it is not. Therefore the Angle BAC is not less than the Angle EDF; but it has likewife been prov'd not to be equal to it. Wherefore the Angle BAC is neceffarily greater than the Angle EDF. If, therefore, two Triangles have two Sides of the one equal to two Sides of the other, each to each, and the Bafe of the one greater than the Base of the other; they fhall also have the Angles, contain❜d under the equal Sides, the one greater than the other; which was to be demonftrated. PROPOSITION XXVI. THEOREM. If two Triangles have two Angles of the one equal to two Angles of the other, each to each, and one Side of the one equal to one Side of the other, either the Side lying between the equal Angles, or which fubtends one of the equal Angles; the remaining Sides of the one Triangle fhall be alfo equal to the remaining Sides of the other, each to his correfpondent Side, and the remaining Angle of the one, equal to the remaining Angle of the other. LET there be two Triangles ABC, DEF, having two Angles ABC, BCA of the one, equal to two Angles DEF, EFD, of the other, each to each, that is, the Angle ABC equal to the Angle DEF and the Angle BCA equal to the Angle EFD. And let one Side of the one be equal to one Side of the other, which first let be the Side lying between the equal Angles, viz. the Side BC equal to the Side EF. I fay, the remaining Sides of the one Triangle will be equal to the remaining Sides of the other, each to each, that is, the Side AB equal to the Side DE, and the Side AC equal to the Side DF, and the remaining Angle BAC equal to the remaining Angle EDF. For For if the Side AB be not equal to the Side DE, one of them will be the greater, which let be AB, make GB equal to DE, and join GC. Then because BG is equal to DE, and BC to EF, the two Sides GB, B C, are equal to the two Sides DE, EF, each to each; and the Angle GBC equal to the Angle DEF. The Bafe GC is equal to the * 4 of this. Bafe DF, and the Triangle G B C to the Triangle DEF, and the remaining Angles equal to the remaining Angles, each to each, which fubtend the equal Sides. Therefore the Angle G C B is equal to the Angle DFE. But the Angle DFE, by the Hypothefis, is equal to the Angle BCA; and fo the Angle BCG is likewife equal to the Angle BCA, the lefs to the greater, which cannot be. Therefore A Bis not unequal to DE, and confequently is equal to it. And fo the two Sides AB, BC, are each equal to the two Sides DE, EF, and the Angle ABC'equal to the Angle D E F And confequently the Bafe AC is equal to the Bafe D F, and the remaining Angle BAC equal to the remaining Angle EDF. Secondly, Let the Sides that are fubtended by the equal Angles be equal, as AB equal to DE. Í fay, the remaining Sides of the one Triangle, are equal to the remaining Sides of the other, viz. A C to DF, and BC to EF; and alfo the remaining Angle BAC, to the remaining Angle EDF. For if BC be unequal to EF, one of them is the greater, which let be BC, if poffible, and make BH equal to EF, and join AH. * Now because BH is equal to EF, and AB to DE, the_two Sides AB, BH, are equal to the two Sides DE, EF, each to each, and they contain equal Angles: Therefore the Base AH is equal to the Base DF; and the Triangle ABH fhall be equal to the Triangle DEF, and the remaining Angles equal to the remaining Angles, each to each, which fubtend the equal Sides: And fo the Angle BHA is equal to the Angle EFD. But EFD is equal to the Angle † From the BCA; and confequently the Angle BHA is equal to Hyp. the Angle BCA: Therefore the outward Angle BHA of the Triangle AHC, is equal to the inward and oppofite Angle BCA; which is impoffible: 16 of this Whence BC is not unequal to EF, therefore it is equal equal to it. But AB is alfo equal to DE. Wherefore the two Sides A B, BC, are equal to the two Sides DE, EF, each to each; and they contain equal Angles. And fo the Bafe AC is equal to the Bafe DF, the Triangle BAC to the Triangle DEF, and the remaining Angle B A C equal to the remaining Angle EDF. If, therefore, two Triangles have two Angles equal, each to each, and one Side of the one equal to one Side of the other, either the Side lying between the equal Angles, or which fubtends one of the equal Angles; the remaining Sides of the one Triangle fall be also equal to the remaining Sides of the other, each to his correspondent Side, and the remaining Angle of the one equal to the remaining Angle of the other; which was to be demonftrated. PROPOSITION XXVII, THEOREM. If a Right Line, falling upon two Right Lines, makes LET the Right Line EF, falling upon two Right For if it be not parallel, AB and CD, produc'd towards B and D, or towards A and C, will meet: Now let them be produc'd towards B and D, and meet in the Point G. * Then the outward Angle AEF of the Triangle *16 of this. GEF, is greater than the inward and oppofite An↑ From the gle EFG, and alfo equal to it; which is abfurd. Hyp. Therefore A B and CD, produc'd towards B and D, will not meet each other. By the fame Way of reafoning, neither will they meet, being produc'd towards C and A. But Lines that meet each other on neiDef. 35. ther Side, are ‡ parallel between themselves. Therefore AB is parallel to CD. Therefore if a Right Line, falling upon two Right Lines, makes the alternate Angles equal between themfelves, the two Right Lines fhall be parallel; which was to be demonftrated. |