4 of this. 6 of this. PROPOSITION IX. THEOREM. Right Lines that are parallel to the fame Right Line, not being in the fame Plane with it, are allo parallel to each other. LET both the Right Lines A B, CD, be parallel to the Right Line EF, not being in the fame Plane with it. I fay A B is parallel to C D. For, affume any Point G in E F, from which Point G. let G H be drawn at Right Angles to E F, in the Plane paffing thro' E F, A B: Alfo, let G K be drawn at Right Angles to E F in the Plane paffing thro' EF, CD: Then, because E F is perpendicular to GH and GK, the Line E F fhall alfo be at Right Angles to a Plane paffing through both GH and GK: But EF is parallel to A B; therefore A B is also at Right Angles to the Plane paffing thro' H G K. For the fame Reason, CD alfo is at Right Angles to the Plane paffing thro' H G K ; and therefore A B, and CD, will be both at Right Angles to the Plane paffing through HGK. But if two Right Lines be at of this. Right Angles to the fame Plane, they shall be* parallel to each other; therefore A B is parallel to CD. And fo, Right Lines that are parallel to the fame Right Line, not being in the fame Plane with it, are alfo parallel to each other; which was to be demonftrated. PROPOSITION X, If two Right Lines, touching one another, be pa- LET two Right Lines A B, BC, touching one another, be parallel to two Right Lines DE, EF, touching one another, but not in the fame Plane. I fay, the Angle A B C is equal to the Angle DEF. Fora For, take B A, BC, ED, EF, equal to one another, and join A D, CF, BE, AC, DF: Then, becaufe B A is equal and parallel to ED, the Right Line A D fhall alfo be equal and parallel to BE. * 33. f. For the fame Reafon, C F will be equal and parallel to BE; therefore A D, CF, are both equal and paralled to BE. But Right Lines that are parallel to the fame Right Line, not being in the fame Plane with it, will be parallel to each other. Therefore A. D† 9 of this. AD is parallel and equal to CF; but A C, D F, join them; wherefore A C is ‡ equal and parallel to DF. And because the two Right Lines A B, BC, are equal to the two Right Lines DE, EF, and the Base A C equal to the Bafe DF; therefore the Angle A B C will be equal to the Angle D E F. Whence, if two Right Lines touching one another, be parallel to two other Right Lines touching one another, but not in the fame Plane, thofe Right Lines contain equal Angles; which was to be demonftrated. PROPOSITION XI. PROBLEM. From a Point given above a Plane, to draw a LETA be the Point given, above the given Plane 33. 1. 8. 1, Let a Right Line BC be any how drawn in the Plane BH; and let AD be drawn * from the Point A,* 12, 1. perpendicular to BC; then if AD be perpendicular to the Plane B H, the Thing required is already done; but, if not, let D E be drawn in the Plane from the Point D, at Right Angles to BC; and let AF be drawn from the Point A, perpendicular to DE: Laftly, through F draw G H, parallel to B C. Then, because B C is perpendicular to both DA and DE, BC will alfo be + perpendicular to a Piane + 4 of tis. paffing thro' E D, DA. But G H is parallel to BC; and if there are two Right Lines parallel, one of which is at Right Angles to fome Plane, then fhall the other be at Right Angles to the fame Plane: 8 Wherefore G H is at Right Angles to the Plane paf fing Def. 3. fing thro' ED, D A, and fo is perpendicular to all the Right Lines, in the fame Plane that touch it. But A F, which is in the Plane paffing thro' E D and DA doth touch it. Therefore, G His perpendicular to A F; and fo AF is perpendicular to G H; but AF likewife is perpendicular to DE; therefore A F is perpendicular to both H G, D E. But if a Right Line ftands at Right Angles to two Right Lines, in +4 of this, their common Section, that Line will be at Right Angles to the Planes paffing through thefe Lines. Therefore AF is perpendicular to the Plane drawn through E D, GH; that is, to the given Plane B Ĥ. Therefore, A F is drawn from the given Point A, above the given Point BH, perpendicular to the faid Plane; which was to be done. * 11 of this, † 31. 1. + 8 of this. PROPOSITION XII. PROBLEM. To erect a Right Line perpendicular to a given L ET A be a given Point in a given Plane MN. It is required to draw a Right Line from the PointA, at Right Angles to the Plane M N. * Let fome Point B be fuppofed above the given Plane, from which let BC be drawn perpendicular to the faid Plane; and let A D be drawn + from A, parallel to B C. Then, because AD, C B, are two parallel Right Lines, one of which, viz. BC is perpendicular to the Plane MN; the other A D, fhall be alfo perpendicular to the fame Plane. Therefore, a Right Line is erected perpendicular to a given Plane, from a Point given therein; which was to be done." PRO PROPOSITION XII. THEORE M. Two Right Lines cannot be erected at Right Angles to a given Plane, from a Point therein given. FOR, if it is poffible, let two Right Lines AB, A C, be erected perpendicular to a given Plane on the fame Side, at a given Point A, in the given Plane. Then let a Plane be drawn thro' BA, A C, cutting the given Plane thro' A in the Right Line * DA E; 3 of thisa but the Line DA E being in the given Plane, touches it; therefore the Right Lines A B, AC, DA E, are in one Plane: And because CA is perpendicular to the given Plane, it fhall also be + perpendicular to all † Def. 3, Right Lines drawn in that Plane, and touching it. Therefore the Angle CA E is a Right Angle. For the fame Reason, BAE is alfo a Right Angle; wherefore the Angle CAE is equal to BA E, and they are both in one Plane; which is abfurd. Therefore, two Right Lines cannot be erected at Right Angles, to a given Plane, from a Point therein given; which was to be demonstrated, PROPOSITION XIV. Thofe Planes, to which the fame Right Line is parallel. For, if they be not, let them be produced 'till they meet each other, and let the Right Line G H be the common Section, in which take any Point K, and join A K, BK. Then, becaufe A B is perpendicular to the Plane EF, it shall also be perpendicular to the Right Line BK, being in the Plane E F produced; wherefore the Plane A B K is a Right Angle. And, for the fame Reason, BAK is also a Right Angle. And 17. I. *11 of this. +31. 1. And fo the two Angles A B K, BAK, of the Triangle A BK, are equal to two Right Angles, which is impoffible: Therefore the Planes C D, E F, being produced, will not meet each other; and fo are neceffarily parallel. Therefore, thofe Planes to which the fame Right Line is perpendicular, are parallel to each other; which was to be demonftrated, PROPOSITION XV. If two Right Lines, touching one another, be pa- LET two Right Lines A B, BC, touching one ano ther, be parallel to two Right Lines DE, EF, touching one another, but not in the fame Plane with them. I fay, the Planes paffing through A B, BC, and DE, EF, being produced, will not meet each other. For, let B G be drawn from the Point B, perpendicular to the Plane paffing through DE, EF, meeting the fame in the Point G; and through G let GH be drawn parallel to ED, and KG parallel to EF; then, because B G is perpendicular to the Plane pafDef. 3. fing through D E, EF, it fhall alfo make* Right Angles with all Right Lines that touch it, and are in the fame Plane. But GH and GK, which are both in the fame Plane, touch it; therefore each of the Angles BGH, BGK, is a Right Angle. And fince BA is parallel to GH, the Angles GBA, BGH, are equal to two Right Angles: But B G H is a Right Angle; wherefore G B A fhall alfo be a Right Angle; and fo BG is perpendicular to B A. For the fame Reason, GB is alfo perpendicular to BC; therefore fince a Right Line BG ftands at Right Angles to two Right Lines B A, B C, mutually cutting 4 of this. each other BG fhall alfo be at Right Angles to the Plane drawn thro' BA, B C. But it is perpendicular to the Plane drawn through DE, EF; therefore #29. I. * BG |