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. *3 of this. Assume any Point Din AC, and make CE equal * to CD, and upon DE make the Equilateral +3 of this, Triangle FDE, and join FC. I say, the Right Line FC is drawn from the Point C, given in the Right Line AB at Right Angles to AB. For because D C is equal to CE, and FC is common, the two Lines DC, CF, are each equal to the two Lines EC, CF; and the Base DF is equal to the Base FE. Therefore * the Angle DCF is equal * 8 of this, to the Angle ECF; and they are adjacent Angles. But when a Right Line, standing upon a Right Line, makes the adjacent Angles equal, each of the equal Angles is a Right Angle; and consequently DCF, ‡ Def. 10. FCE, are both Right Angles. Therefore the Right Line FC, &c. which was to be done. PROPOSITION ΧΙΙ. PROBLEM. To draw a Right Line perpendicular, upon a given infinite Right Line, from a Point given out of it. L 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. Assume any Point D on the other Side of the Right Line AB, and about the Center C, with the Distance CD describe * a Circle EDG, bisect + EG in H, * Pol. 3.. and join CG, CH, CE. I say there is drawn the † 10 of this. Per 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 $ 8 of this, the Angle CHG is equal ‡ to the Angle CHE; and they are adjacent Angles. But when a Right Line, standing 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 stands on. Therefore CH is drawn perpendicular, upon a given infinite Right Line, from a given Point out of it; which was to be demonftrated. PROPOSITION XIII. THEOREM. When a Right Line, standing upon a Right Line, makes Angles, these shall be either two Right Angles, or together equal to two Right Angles. F OR let a Right Line A B, standing upon the Right Line CD, make the Angles CBA, A B D. I fay, the Angles CBA, ABD, are either two Right Angles, or both together equal to two Right Angles. * Def. &. Ax. 2. 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 EBD, 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 DBA, ABC, are equal * Ax 1. to the three Angles DBE, EBA, ABC, together. But it has been prov'd, that the two Angles CВЕ, EBD, together, are likewise equal to these three Angles: But Things that are equal to one and the same, are * equal between themselves. Therefore likewiso the Angles CBE, EBD, together, are equal to the 2 Angles 1 Angles DBA, ABC, together; but CBE, EBD, are two Right Angles. Therefore the Angles D BA, ABC, are both together equal to two Right Angles. Wherefore when a Right Line, standing upon another Right Line, makes Angles, these shall be either two Right Angles, or together equal to two Right Angles; which was to be demonstrated. 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 said two Right Lines will make but one straight Line. FOR COR 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 say, BC, BD, make but one Right Line. For if BD, CB, do not make one straight Line, let CB and BE make one. 1 Then, because the Right Line A B stands upon the Right Line CBE, the Angles ABC, ABE, 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 less to the greater, which is impoffible. Therefore BE, BC, are not one straight Line. And in the same Manner it is demonstrated, that no other Line but BD is in a straight Line with CB; wherefore CB, BD, shall 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 Straight Line; which was to be demonstrated. PRO PROPOSITION XV. THEOREM. If two Right Lines mutually cut each other, the opposite Angles are equal. L ET the two Right Lines A B, CD mutually cut each other in the Point E. I fay, the Angle AEC is equal to the Angle DEB; and the Angle CЕВ equal to the Angle A ED. For because the Right Line AE, standing on the Right Line CD, makes the Angles CEA, AED: * 13 of this. These both together shall be equal * to two Right Angles. Again, because the Right Line DE standing upon the Right Line AB, makes the Angles AED, DEB: These Angles together are * equal to two Right Angles. But it has been prov'd, that the Angles CEA, AED, are likewise together equal to two Right Angles. Therefore the Angles CEA, AED, are equal to the Angles A ED, DEB. Take away the common Angle AED, and the Angle remaining CEA, is † equal to the Angle remaining BED. For the same Reason, the Angle CEB shall be equal to the Angle DEA. Therefore if two Right Lines mutually cut each other, the opposite Angles are equal; which was to be demonftrated. † Ax. 3. Coroll. 1. From hence it is manifest, that two Right 1 PROPROPOSITION XVI. THEOREM. If one Side of any Triangle be produced, the outward Angle is greater than either of the inward opposite Angles. L ET ABC be a Triangle, and one of its Sides BC, be produced to D. I say, the outward Angle ACD is greater than either of the inward Angles CBA, or BAC. For bisect AC in E*, and join BE, which pro- * IO of this. duce to F, and make EF equal to BE. join FC, and produce AC to G. Moreover, 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 AEB tequal to the Angle FEC; for they are opposite † 15 of this. Angles. Therefore the Base AB, is ‡ equal to the 14 of this. Base FC; and the Triangle AE 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 ECD, 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 bisected, we demonftrate 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. THEOREM. Two Angles of any Triangle together, howsoever LET Right Angles. For |
