PROPOSITION IX. PROBLEM. To cut a given Right-lin❜d Angle into two equal Parts. LET BAC be a given Right-lin❜d Angle, which is required to be cut into two equal Parts. Affume any Point D in the Right Line A B, and + 3 of this. cut off A E from the Line A C equal to AD+; join 11 of this. DE, and thereon make the equilateral Triangle DEF, and join A F. I fay, the Angle B A C is cut into two equal Parts by the Line A F. For, because A D is equal to A E, and A F is common, the two Sides D'A, A F, are equal to the two Sides A E, A F, and the Bafe D F is equal to the 8 of this. Bafe E F; therefore I the Angle D A F is equal to the Angle E A F. Wherefore, a given Right-lin’d Angle is cut into two equal Parts; which was to be done. I of this. + 9 of this. PROPOSITION X.. PROBLEM. To cut a given finite Right Line into two equal LET AB be a given finite Right Line, required to be cut into two equal Parts. Upon it make an Equilateral Triangle A B C, and bilect + the Angle A CB by the Right Line CD. I fay the Right Line A B is bifected in the Point D. For, because AC is equal to C B, and C D is common, the Right Lines A C, CD, are equal to the two Right Lines BC, CD, and the Angle ACD #4 of this. equal to the Angle B CD; therefore I the Bafe A D is equal to the Bafe DB. And fo, the Right Line AB is bifected in the Point D; which was to be done. PRO PROPOSITION XI. 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 C'E equal to CD; and upon D E make + the Equilateral 3 of this. Triangle F D E, and join FC. I fay, the Right † 1 of this. Line FC is drawn from the Point C, given in the Right Line A B, at Right Angles to A B. For, because DC is equal to C E, and F C is common, the two Lines D C, CF, are equal to the two Lines EC, CF; and the Bafe DF is equal to the Base F E. Therefore the Angle DC F 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 D C F, † 10 of this. FCE, are both Right Angles. Therefore, the Right Line FC is drawn from the Point Cat Right Angles to A B; which was to be done. PR POPSITION 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 + E G in H, Poft. 3. and join CG, CH, CE. I fay, there is drawn the † 10 of this.* Per Perpendicular C H on the given infinite Right Line For, becaufe G H is equal to H E, and HC is common, G H and HC are each equal to E H and HC, and the Bafe CG is equal to the Bafe C E. Therefore ↑ 8 of this. the Angle C H G 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 themfelves, each of the equal Angles Def. 1o. is a Right one, and the faid ftanding Right Line is called 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; Ax. 2. which was to be done. PROPOSITION XIII. THRORE M. 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 let a Right Line A B, ftanding upon the Right Line CD, make the Angles C B A, ABD. I fay, the Angles CBA, ABD, are either two Right Angles, or both together equal to two Right Angles. * * De 10. For if C B A be equal to A B D, they are each of +11 of this. them Right Angles: But if not, draw + B E 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 the two Angles CBE, E B D, 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 A B C, and the two Angles D BA, A B C, are equal to the three Angles D BE, E BA, ABC, together. But it has been proved, that the two Angles CBE, EBD, together, are likewife equal to thefe three Angles: But Things that are equal to one and * the fame, are equal between themfelves. Therefore like wife the Angle CBE, EBD, together, are equal Ax. 1. to to the Angle B DA, 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, 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 ftrait Line. FOR let two Right Lines B C, BD, drawn from contrary Parts to the Point B, in any Right Line AB, make the adjacent Angles A BC, A BD, both together equal to two Right Angles. I fay, BC, BD, make but one Right Line. For if B D, C B, 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 CBE, the Angle ABC, ABE, together, will be equal to two Right Angles. But the Angles 13 of bin. ABC, ABD, together are alfo equal to two Right Angles. Now taking away the common Angle ABC, the remaining Angle A B E 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 demonftrated, that no other Line but B D is in a ftrait Line with CB; wherefore C B, B D, fhall be in one ftrait 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 ftrait Line; which was to be demonftrated. PROPOSITION XV. THEORE M. If two Right Lines mutually cut each other, the LET the two Right Lines A B, CD, mutually cut For because the Right Line A E, ftanding on the Right Line C D, makes the Angles CEA, A ED; 13 of this. thefe both together fhall be equal to two Right Angles. Again, because the Right Line D E, standing 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 CEA, A E D, are equal to the Angles AED, DEB. Take away the common Angle A ED, and the Angle remaining C E A is +Ax. 2. 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 oppofue Angles are equal; which was to be demonftrated. Coroll. 1. From hence it is manifeft, that two Right PRO |