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 Base DF, the Triangle BAC to the Triangle DEF, and the remaining Angle BAC equal to the remaining Angle EDF. If, therefore, two Triangles have twa 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 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 which was to be demonstrated. 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 EF, falling upon two Right Lines AB, CD, make the alternate Angles AEF, EFD, equal between themselves. I fay the Right Line AB is parallel to CD. For if it be not parallel, AB 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. 1 Then the outward Angle AEF of the Triangle 16 of this. GEF, is greater than the inward and oppofite AnFrom the gle EFG, and alfo equal + to it; which is abfurd. Hyp. † Def. 35. Therefore AB and CD, produced towards B and D, 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 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 demonstrated. PRO PROPOSITION XXVIII. THEOREM. If a Right Line, falling upon two Right Lines, makes the outward Angle of the one Line equal to the inward and oppofite Angle of 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. ET the Right Line EF falling upon two Right Lines AB, 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 AB is parallel to the Right Line CD. 115 of this. For because the Angle EGB is equal to the An- * From the gle GHD, and the Angle EGB + equal to the An- Hyp. gle AGH, the Angle AGH fhall be equal to the Angle GHD; but these are alternate Angles. Therefore AB is parallel to CD. * 27 of this. Again, because the Angles BGH, GHD, are equal to two Right Angles, and AGH, BGH, are equal* 13 of this. to two Right Ones, the Angles AGH, 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 these are alternate Angles. Therefore AB is parallel to CD. If, therefore, a Right Line, falling upon two Right Lines, makes the outward Angle of the one Line equal to the inward and oppofite Angle of 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. THEORE M. 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. 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 GB, 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 GHD, one of them will be the greater. Let this be AGH; then because the Angle AGH is greater than the Angle GHD, add the common Angle BGH to both: And fo the Angles AGH, BGH together, are greater than the Angles BGH, GHD, together. But the Angles * 13 of this. AGH, BGH, are equal to two Right ones *. Therefore BGH, GHD, are lefs than two Right Angles. 4x. 12. And fo the Lines AB, CD, infinitely produced t, will meet each other; but because they are parallel, they will not meet. Therefore the Angle AGH is not unequal to the Angle GHD. Wherefore it is ne ceffarily equal to it. 1.25 of this. But the Angle AGH ist equal to the Angle EGB: Therefore EGB is alfo equal to GHD Now add the common Angle BGH, and then EGB, BGH, together, are equal to BGH, GHD, together; but EGB, and BGH, are equal to two Right Angles. Therefore allo 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. THE ORE M. Right Lines parallel to one and the fame Right LET AB and CD be Right Lines, each of which Then because the Right Line GK falls upon the parallel Right Lines AB, EF, the Angle AGH is* equal * 29 of thite to the Angle GHF; and because the Right Line GK, falls upon the parallel Right Lines EF, CD, the Angle GHF is equal to the Angle GKD*. But it has been proved, that the Angle AGK is alfo equal to the Angle GHF. Therefore AGK is equal to GKD and they are alternate Angles, whence AB is parallel to CDt. And fo Right Lines parallel to one and the † 27 of this fame Right Line, are parallel between themselves; which was to be demonftrated. PROPOSITION XXXI. PROBLEM. To draw a Right Line thro' a given Point parallel to a given Right Line. LE ET A be a Point given, and BC a Right Line the Point A, parallel to the Right Line BC. Affume any Point D in BC, and join AD; then make* an Angle DAE, at the Point A, with the * 23 of this Line DA, equal to the Angle ADC, and produce EA ftrait forwards to F. Then because the Right Line AD falling on two Right Lines BC, EF, makes the alternate Angles EAD, ADC, equal between themselves, EF fhall bet parallel to BC. Therefore the Right Line EAF † 27 of this. is drawn thro' the given Point A, parallel to the given Right Line BC; which was to be done. Coroll. * 31 of this. Coroll. Hence it appears, that if one Angle of any Triangle be equal to the other two, that is a Right one; because that the Angle adjacent to this Right one, is equal to the other two. But when adjacent Angles are equal, they are neceffarily Right ones. PROPOSITION. XXXII. THEOREM. If one Side of any Triangle be produced, the outward LET ABC be a Triangle, one of whofe Sides BC is produced to D. I fay, the outward Angle ACD is equal to the two inward and oppofite Angles CAB, ABC; and the three inward Angles of the Triangle, viz. ABC, BCA, CAB, are equal to two Right Angles. For let CE be drawn thro' the Point C parallel to the Right Line AB. Then because AB is parallel to CE, and AC falls upon them, the alternate Angles t29 of this. BAC, ACE, are † equal between themselves. Again, because AB is parallel to CE, and the Right Line BD falls upon them, the outward Angle ECD is t equal to the inward and oppofite one ABC; but it has been proved, that the Angle ACE is equal to the Angle BAC. Wherefore the whole outward Angle ACD is equal to both the inward and oppofite Angles BAC, ABC. And if the Angle ACB, which is common, be added, the two Angles ACD, ACB, are equal to the three Angles ABC, BAC, 13 of this. ACB; but the Angles ACD, ACB, are equal to two Right Angles. Therefore also fhall the Angles ACB, CBA, CAB, be equal to two Right Angles. Wherefore, if one Side of any Triangle be produced, the outward Angle is equal to both the inward and oppofite Angles, and the three inward Angles of a Triangle are equal to two Right Angles which was to be demonftrated. Corall |