PROPOSITION XI. PROBLEM. To draw a Right Line at Right Angles to a given L' ET AB be the given Right Line, and C the given Point. It is required to draw a Right Line from the Point C, at Right Angles to AB. * Affume any Point D in AC, 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 AB. For because DC is equal to CE, and F C is common, the two Lines DC, CF, are each equal to the two Lines EC, CF; and the Bafe DF is equal to the Bafe FE. 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, ‡ Def. 10. FCE, are both Right Angles. Therefore the Right Line FC, &c. which was to be done. PROPOSITION XII. PROBLEM. To draw a Right Line perpendicular, upon a given infinite Right Line, from a Point given out of it. ET AB be the given infinite Line, and C the Point given out of it. It is requir'd to draw a Right Line perpendicular upon the given Right Line AB, from the Point C given out of it. Affume any Point D on the other Side of the Right Line A B, and about the Center C, with the Distance CD describe a Circle EDG, bifect + EG in H, and join CG, CH, CE. I fay there is drawn the Per Poff. 3.. † 10 of this. Perpendicular CH on the given infinite Right Line For because G H is equal to HE, and HC is common, GH and HC are each equal to EH and HC, and the Base CG is equal to the Base 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 said standing 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 demonstrated. + Ax. 2. 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. OR let a Right Line A B, ftanding upon the Right FOR AB, Line CD, make the Angles CBA, ABD. I say, the Angles CBA, ABD, are either two Right Angles, or both together equal to two Right Angles. * *Def. 8. For if CBA be equal to ABD, 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 BD, which is common; and the two Angles CBE, EBD, together, are equal to the three Angles CBA, ABE, EBD, together. Again, because the Angle DBA is equal to the two Angles DBE, EBA, together, add the common Angle ABC, and the two Angles D BA, ABC, are equal to the three Angles DBE, EBA, ABC, together. But it has been prov'd, that the two Angles CBE, EBD, together, are likewife equal to these three Angles But Things that are equal to one and the fame, are equal between themselves. Therefore likewife the Angles CBE, EBD, together, are equal to the * Ax 1. * Angles Angles DBA, A B C, together; but CBE, EBD, are two Right Angles. Therefore the Angles DBA, ABC, are both together equal to two Right Angles. Wherefore when a Right Line, ftanding 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 ftraight Line. FOR OR 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 BD, CB, do not make one straight Line, let CB and BE make one. Then, because the Right Line A B ftands upon the Right Line CBE, the Angles ABC, AB 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, and 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 fstraight Line. And in the fame Manner it is demonftrated, that no other Line but BD is in a straight Line with CB; wherefore CB, BD, fhall be in one straight 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 two Right Angles, the faid two Right Lines will make but one ftraight Line; which was to be demonstrated. PRO PROPOSITION XV. THEOREM. If two Right Lines mutually cut each other, the oppofite Angles are equal. LET the two Right Lines AB, 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 Ă ED. * For because the Right Line A E, ftanding on the Right Line CD, makes the Angles CEA, AED: 13 of this. These both together fhall be equal to two Right Angles. Again, because the Right Line DE ftanding upon the Right Line AB, makes the Angles AED, DEB: Thefe Angles together are* equal to two Right Angles. But it has been prov'd, that the Angles CEA, AED, are likewife together equal to two Right Angles. Therefore the Angles CEA, AED, are equal to the Angles AED, DEB. Take away the common Angle AED, and the Angle remaining CEA, is t equal to the Angle 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 Coroll. 2. All the Angles conftituted about the fame PRO PROPOSITION XVI. THEORE M. If one Side of any Triangle be produced, the outward Angle is greater than either of the inward oppofite Angles. LET ET ABC be a Triangle, and one of its Sides BC, be produced to D. I fay, the outward Angle ACD is greater than either of the inward Angles CBA, or BA C. For bifect AC in E*, and join BE, which pro- * 10 of this duce to F, and make EF equal to BE. Moreover, join FC, and produce A C to G. Then, because AE is equal to EC, and BE to EF, the two Sides AE, EB, are equal to the two Sides CE, EF, each to each, and the Angle A EB tequal to the Angle FEC; for they are oppofite † 15 of this. Angles. Therefore the Base A B, is equal to the ‡ 4 of this. Base FC; and the Triangle A E B, equal to the Triangle FEC; 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 BAE, is equal to the Angle ECF; but the Angle E CD, is greater than the Angle ECF; therefore the Angle ACD, is greater than the Angle BAE. After the fame manner, if the Right Line BC, be bifected, we demonstrate that the Angle BCG, that is, the Angle A CD, is greater than the Angle ABC. Therefore one Side of any Triangle being produced, the outward Angle is greater than either of the inward oppofite Angles; which was to be demonftrated. PROPOSITION XVII. THEORE M. Two Angles of any Triangle together, bowfoever taken, are less than two Right Angles. L ET ABC be a Triangle. I fay, two Angles of it together, howfoever taken, are less than two Right Angles, For |