PROPOSITION ΧΙ. PROBLEM. To draw a Right Line at Right Angles to a given LETAB be the given Right Line, and C the given the Point C, at Right Angles to A B. Affume any Point D in A C, and make CE equal to CD; and upon DE make + the Equilateral Triangle FDE, and join FC. I fay, the Right Line FC is drawn from the Point C, given in the Right Line A B, at Right Angles to A B. * 3 of this. of this. For, because DC is equal to C E, and F C is common, the two Lines DC, CF, are equal to the two Lines E C, CF; and the Bafe D F is equal to the Base F E. Therefore the Angle DCF is equal 8 of this. to the Angle ECF; and they are adjacent Angles. But when a Right Line, ftanding upon a Right Line, makes the adjacent Angles equal, each of the equal Angles is a Right Angle; and confequently DCF, 10 of this. FCE, are both Right Angles. Therefore, the Right Line FC is drawn from the Point C at Right Angles to AB; which was to be done. PROPOSITION XII. PROBLEM. To draw a Right Line perpendicular, upon a given LET AB be the given infinite Line, and C the Affume any Point D on the other Side of the Right CD, defcribe a Circle E DG; bifect + EG in H, P. 3. t and join CG, CH, CE. I fay, there is drawn the † 10 of this. Per Perpendicular CH on the given infinite Right Line For, because G H is equal to H E, and HC is common, GH and HC are each equal to E H and HC, and the Base CG is equal to the Bafe CE. Therefore 18 of this, the Angle CHG is equal to the Angle CHE; and they are adjacent Angles. But when a Right Line, ftanding upon another Right Line, makes the Angles equal between themselves, each of the equal Angles Def. 10. is a Right one, and the faid ftanding Right Line is call'd a Perpendicular to that which it ftands on. Therefore, CH is drawn perpendicular, upon a given infinite Right Line, from a given Point out of it; which was to be done. PROPOSITION XIII. THEOREM. When a Right Line, ftanding upon a Right Line, makes Angles, thefe fhall be either two Right Angles, or together equal to two Right Angles. FOR OR let a Right Line A B, ftanding upon the Right Line CD, make the Angles CBA, ABD. Hay, the Angles C B A, ABD, are either two Right Angles, or both together equal to two Right Angles. * Def. 10. For if CB A be equal to A B D, they are each of +11 of this. them Right Angles: But if not, draw + BE from the Point B, at Right Angles to CD. Therefore the Angles CBE, EBD, are two Right Angles: And because CBE is equal to both the Angles CBA, ABE, add the Angle E B D, which is common; and Ax. 2. the two Angles CBE, EBD, together, are equal to the three Angles CBA, A BE, EBD, together. Again, because the Angle DBA is equal to the two. Angles D BE, E B A, together, add the common Angle ABC, and the two Angles DBA, ABC, are equal to the three Angles DBE, EBA, ABC, together. But it has been proved, that the two Angles CBE, EBD, together, are likewise equal to these three Angles But Things that are equal to one and the fame, are * equal between themselves. Therefo e likewife the Angle CBE, EBD, together, are equal * Ax. I. : to the Angles DBA, ABC, together; but CBE, EBD, are two Right Angles. Therefore the Angles DBA, A B C, are both together equal to two Right Angles. Wherefore, when a Right Line, flanding upon another Right Line, makes Angles, thefe fhall be either two Right Angles, or together equal to two Right Angles; which was to be demonftrated. PROPOSITION XIV. THEOREM. If to any Right Line, and Point therein, two Right Lines be drawn from contrary Parts, making the adjacent Angles, both together, equal to two Right Angles, the faid two Right Lines will make but one ftrait Line. FOR let two Right Lines BC, BD, drawn from contrary Parts to the Point B, in any Right Line A B, make the adjacent Angles ABC, ABD, both together equal to two Right Angles. I fay, BC, BD, make but one Right Line. For if B.D, CB, do not make one ftrait Line, let CB and B E make one. * * Then, because the Right Line A B ftands upon the Right Line C B E, the Angle A B C, A B E, together, will be equal to two Right Angles. But the Angles 13 of this. ABC, ABD, together, are also equal to two Right Angles. Now taking away the common Angle ABC, the remaining Angle ABE is equal to the remaining Angle ABD, the lefs to the greater, which is impoffible. Therefore BE, BC, are not one ftrait Line. And in the fame manner it is demonstrated, that no other Line but BD is in a ftrait Line with CB; wherefore CB, BD, fhall be in one strait Line. Therefore, if to any Right Line, and Point therein, two Right Lines be drawn from contrary Parts, making the adjacent Angles, both together, equal to twe Right Angles, the faid two Right Lines will make but one frait Line; which was to be demonstrated. PRO 1 PROPOSITION XV. THEOREM. If two Right Lines mutually cut each other, the oppofite Angles are equal. ET the two Right Lines A B, CD, mutually cut each other in the Point E. I say, the Angle AEC is equal to the Angle DEB; and the Angle CEB equal to the Angle A E D. * For, because the Right Line A E, ftanding on the Right Line CD, makes the Angles CEA, AED; 13 of bis.these both together fhall be equal to two Right Angles. Again, because the Right Line D E, ftanding upon the Right Line A B, makes the Angles A E D, DEB; thefe Angles together are equal to two Right Angles. But it has been proved, that the Angles CEA, AED, are likewife together equal to two Right Angles. Therefore the Angles CE A, A E D, are equal to the Angles AED, DEB. Take away the common Angle A E D, and the Angle remaining CEA is † equal to the Angie remaining BED. For the fame Reason, the Angle CEB fhall be equal to the Angle DEA. Therefore, if two Right Lines mutually cut each other, the oppofite Angles are equal, which was to be demonftrated. † Ax. 3. Coroll. 1. From hence it is manifeft, that two Right PRO PROPOSITION XVI. THEOREM. If one Side of any Triangle be produced, the out- LET ABC be a Triangle, and one of its Sides For bifect AC in E*, and join BE, which pro- 100f thita. duce to F, and make E F equal to BE+. Moreover, † 3 of ibis. join F C, and produce A C to G. Then, because AE is equal to EC, and B E to EF, the two Sides A E, E B, are equal to the two Sides CE, EF, each to each, and the Angle AEB equal to the Angle FEC; for they are oppofite † 15 of this. Angles. Therefore the Bafe AB is equal to the 14 of this. Bale FC; and the Triangle A E B equal to the Triangle F E C; and the remaining Angles of the one equal to the remaining Angles of the other, each to each, fubtending the equal Sides. Wherefore the Angle B A E is equal to the Angle ECF; but the Angle ECD is greater than the Angle ECF; therefore the Angle A CD is greater than the Angle B A E. After the fame manner, if the Right Line BC be bifected, we demonftrate that the Angle BCG, and confequently its equal, the Angle A CD*, is greater 15 of ibis, than the Angle ABC. Therefore, one Side of any Triangle being produced, the outward Angle is greater than either of the inward appofite Angles; which was to be demonftrated. PROPOSITION XVII. THEOREM. Two Angles of any Triangle together, howsoever taken, are less than two Right Angles. ET ABC be a Triangle. I fay, two Anglesof it together, howsoever taken, are less than two Right Angles. For |