thefis, is equal to the Angle BC A; and fo the Angle BCG is likewife equal to the Angle BCA, the less to the greater, which cannot be. Therefore A B is not unequal to D E, and confequently is equal to it. And fo the two Sides A B, BC, are equal to the two Sides DE, EF, and the Angle ABC equal to the 4 of this. Angle DEF: And confequently the Bafe AC is equal to the Bafe DF, and the remaining Angle BAC equal to the remaining Angle ED F. Secondly, Let the Sides that are fubtended by the equal Angles be equal, as A B equal to D E. Í fay, the remaining Sides of the one Triangle are equal to the remaining Sides of the other, viz. AC to DF, and BC to EF; and alfo the remaining Angle BAC, to the remaining Angle E DF. For if BC be unequal to EF, one of them is the greater, which let be B C, if poffible, and make BH equal to E F, and join A H. Now, becaufe BH is equal to EF, and A B to DE, the two Sides A B, BH, are equal to the two Sides DE, EF, each to each, and they contain equal Angles: Therefore the Bafe AH is + equal to the Bale DF; and the Triangle A B H fhall be equal to the Triangle D E F, 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 From the the Angle EF D. But EFD is + equal to the AnHyp gle BCA; and confequently the Angle BHA is equal to the Angle BCA: Therefore the outward Angle B HA of the Triangle A HC, is equal to the 16 of this. inward and oppofite Angle BCA; which is impoffible: Whence B C is not unequal to E F ; there, fore it is equal to it. But A B is alfo equal to DE. Wherefore the two Sides A B, BC, are equal to the two Sides DE, E F, each to each; and they contain equal Angles. And fo the Bafe AC is equal to the Bafe D F, the Triangle BAC to the Triangle DEF, and the remaining Angle B A C equal to the remain, ing Angle EDF. If, therefore, twa 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 ane of the equal Angles; the remaining Sides of the one Triangle shall be alfa equal to the remaining Sides of the other, each to its corre Spondent Spondent 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 the alternate Angles equal between themfelves, the two Right Lines fhall be parallel. L ET the Right Line E F, falling upon two Right Lines AB, CD, make the alternate Angles AEF, EFD, equal between themselves. I fay, the Right Line A B is parallel to C D. For if it be not parallel, A B and CD, produced towards B and D, or towards A and C, will meet : Now let them be produced towards B and D, and meet in the Point G. Then the outward Angle A EF of the Triangle GEF is greater than the inward and oppofite An- 16 of this gle EFG, and alfo equal + to it; which is abfurd. + From the Therefore A B and C D, produced towards B and D, Hyp will not meet each other. By the fame Way of Reafoning, neither will they meet, being produced towards C and A. But Lines that meet each other on neither Side, are parallel between themselves. Therefore ‡¡Def. 35 A B 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. PRO * From the Hyp. PROPOSITION XXVII. THE ORE M. If a Right Line, falling upon two Right Lines, makes the outward Angle with the one Line equal to the outward and oppofite Angle with the other on the fame Side, or the inward Angles on the fame Side together equal to two Right Angles, the two Right Lines fhall be parallel between themselves. L ET the Right Line EF falling upon two Right Lines A B, CD, make the outward Angle EGB equal to the inward and oppofite Angle GHD; or the inward Angles BGH, GHD, on the fame Side together equal to two Right Angles. I fay, the Right Line A B is parallel to the Right Line CD. * For, because the Angle EGB is equal to the An15 of this.gle GHD, and the Angle EGB+ equal to the Angle AGH, the Angle AG H fhall be equal to the Angle GHD; but thefe are alternate Angles. 127 of this. Therefore A B is parallel to CD. Again, because the Angles BGH, GHD, are 13 of this. equal to two Right Angles, and AGH, BGH, are equal to two Right ones, the Angles AG H, BGH, will be equal to the Angles BGH, GHD; and if the common Angle BGH be taken from both, there will remain the Angle AGH equal to the Angle GHD; but thefe are alternate Angles. Therefore A B is parallel to CD. If, therefore, a Right Line, falling upon two Right Lines, makes the outward Angle with the one Line equal to the inward and oppofite Angle with the other on the fame Side, or the inward Angles on the fame Side together equal to two Right Angles, the two Right Lines fhall be parallel between themselves; which was to be demonftrated. PRO. PROPOSITION XXIX. T.HEOREM. If a Right Line falls upon two Parallels, it will make the alternate Angles equal between themfelves; the outward Angle equal to the inward and oppofite Angle, on the fame Side; and the inward Angles on the fame Side together equal to two Right Angles. LET ET the Right Line EF fall upon the parallel Right Lines AB, CD. I fay, the alternate Angles AGH, GHD, are equal between themselves; the outward Angle E G B is equal to the inward one GHD, on the fame Side; and the two inward ones, BGH, GHD, on the fame Side, are together equal to two Right Angles. For if AGH be unequal to G HD, one of them will be the greater. Let this be A GH; then because the Angle A G H is greater than the Angle G H D, add the common Angle B G H to both: And fo the Angles AGH, BGH, together, are greater than the Angles BGH, GHD, together. But the Angles AGH, BGH, are equal to two Right ones *. * 13 of this, Therefore BGH, GHD, are lefs than two Right Angles. And fo the Lines A B, CD, infinitely produced + will meet each other; but because they are† Ax. 12. parallel they will not meet. Therefore the Angle AGH is not unequal to the Angle GHD. Wherefore it is neceffarily equal to it. But the Angle AGH is equal to the Angle ‡ 15 of this. EGB: Therefore E G B is alfo equal to GHD. Now add the common Angle BGH; and then EGB, BGH, together, are equal to BG H, GHD, together; but EGB, and BG H, are equal to two Right Angles. Therefore alfo BGH, and GHD, fhall be equal to two Right Angles. Wherefore, if a Right Line falls upon two Parallels, it will make the alternate Angles equal between themselves; the outward Angle equal to the inward and oppofite Angle, on the fame Side; and the inward Angles on the fame Side together equal to two Right Angles; which was to be demonftrated. PRO PROPOSITION XXX. THEOREM. Right Lines, parallel to one and the fame Right LET ET AB and CD be Right Lines, each of which is parallel to the Right Line EF. I fay, A B is alfo parallel to CD. For let the Right Line GK fall upon them. Then, because the Right Line GK falls upon the * 29 of this, parallel Right Lines AB, EF, the Angle AGH is * equal to the Angle GHF; and because the Right Line GK falls upon the parallel Right Lines EF, CD, † 29 of bit. the Angle GHF is equal to the Angle G K D†. But it has been proved that the Angle AG K is alfo equal to the Angle GHF. Therefore A G K is equal to GKD, and they are alternate Angles; whence A B +27 of this, is parallel to CD+. And fo, Right Lines, parallel to one and the fame Right Line, are parallel between themfelves; which was to be demonftrated. PROPOSITION XXXI. PROBLEM. To draw a Right Line through a given Point parallel to a given Right Line. LE ET A be a Point given, and BC a Right Line given. It is required to draw a Right Line thro' the Point A, parallel to the Right Line B C. Affume any Point D in BC, and join AD; then 123 of this, make an Angle D A E, at the Point A, with the Line DA, equal to the Angle ADC, and produce EA ftrait forwards to F. Then, because the Right Line A D, falling on two Right Lines BC, EF, makes the alternate Angles EAD, ADC, equal between themselves, EF fhall +27 of this. bet parallel to BC. Therefore, the Right Line EAF is drawn thro' the given Point A, parallel to the given Right Line BC; which was to be done. PRO 1 |