PROPOSITION IX. THEOREM. Line, not being in the same Plane with it, are also parallel 10 each other. LET both the Right Lines AB, CD, be parallel to the Right Line EF, not being in the same Plane with it. I say A B is parallel to CD. For, afsume 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' EF, AB: Allo, let G K be drawn ar Right Angles to E F in the Plane paffing thro’EF, CD: Then, because E F is perpendicular to GH 4 of tbis. and G K, the Line E F shall allo be * at Right Angles to a Plane pafling through both G Hand GK: But 6 of tbis. E F is parallel to A B; therefore A B is + also at Right Angles to the Plane paffing thro' HGK. For fing through HGK. Bur if iwo Right Lines be at 6 of tbis. Right Angles to the same Plane, they ihall be * pa rallel to each other; therefore A B is parallel to CD. And so, Right Lines that are parallel to the same Right Line, not being in the same Plane with it, are also parallel to each other ; which was to be demonstrated. PROPOSITION X, THEOREM. rallel to two other Right Lines, touching one Lines contain equal Angles. ther, be parallel to two Right Lines D E, EF, touching one another, but not in the fame Plane. I say, the Angle A B C is equal to the Angle DEF. For, + For, take B A, BC, ED, EF, equal to one another, and join A D, CF, BE, AC, DF: Then, because B A is equal and parallel to ED, the Right Line A D shall also be * equal and parallel to B E. * 33. $. For the same Reason, C F will be equal and parallel to B E; therefore AD, CF, are both equal and paralled to B E. But Right Lines that are parallel to the same Right Line, not being in the same Plane with it, will be + parallel to each other. Therefore A Dt of tbis, is parallel and equal to CF; but A C, D F, join them; wherefore A Cis I equal and parallel to DF. I 33. 1. And because the two Right Lines AB, BC, are equal to the two Right Lines DE, EF, and the Base AC equal to the Base DF; therefore the Angle A BC will be * equal to the Angle D EF. Whence, if two 8. 1, Right Lines touching one unother, be parallel to two other Right Lines touching one another, but not in the same Plane, those Right Lines contain equal Angles; which was to be demonstrated. PROPOSITION XI. PROBLEM. Right Line perpendicular to that Plane. BH. It is required to draw a Right Line from the Point A, perpendicular to the Plane B H, Let a Right Line BC be any how drawn in the Plane BH; and let S D be drawn * from the Point A, 12, 1. perpendicular to BC; then if A D be perpendicular to the Plane B H, the Thing required is already done; þut, if not, let D E be drawn in the Plane from the Point D, at Right Angles to B C; 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 D A and D E, B C will also be + perpendicular to a Piane + 4 of tis. passing thro' E D, D A. But G H is parallel to BC; and if there are two Right Lines parallel, one of which is at Right Angles to some Plane, then shall the other be I at Right Angles to the fame Plane : 8 sikit, Wherefore G H is at Right Angles to the Plane pal • Def. 3. fing thro' E D, D A, and so is * perpendicular to all the Right Lines, in the same Plane that touch it. But A F, which is in the Plane paffing thro' E D and D A doth touch it. Therefore, G His perpendicular to AF; and so A F is perpendicular to GH; but AF likewise is perpendicular to DE; therefore A F is perpendicular to both HG, DE. But if a Right Line stands at Right Angles to two Right Lines, in + 4 of ibis, their common Section, that Line will be † at Right Angles to the Planes paffing through these Lines. Therefore AF is perpendicular to the Plane drawn through E D, GH; that is, to the given Plane B H. Therefore, A F is drawn from the given Point A, above the given Point B H, perpendicular to the said Plane ; which was to be done. PROPOSITION XII. PROBLEM. To erezt a Right Line perpendicular to a given Plane, from a Point given therein. LET A be a given Point in a given Plane MN. It is required to draw a Right Line from the Point A, at Right Angles to the Plane M N. Let some point B be supposed above the given . 11 of tbis. Plane, from which let B C be drawn * perpendicular + 31. 1. to the said 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. B C is perpendicular to the + 8 of this. Plane MN; the other A D, shall be fallo perpen dicular 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 XVI. THEOREM. gles to a given Plane, from a Point therein given. FOR, if it is possible, let?two Right Lines A B, 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' B.A, A C, cutting the given Plane thro' A in the Right Line * DAE; •3 of tbisa but the Line D A E being in the given Plane, touches it; therefore the Right Lines A B, AC, DAE, are in one Plane: And because CA is perpendicular to the given Plane, it shall also be + perpendicular to all + Def. 3. Right Lines drawn in that Plane, and touching it. Therefore the Angle CAE is a Right Angle. For the same Reason, B.AE is also a Right Angle ; wherefore the Angle CAE is equal to B A E, and they are both in one Plane ; which is absurd. Therefore, two Right Lines cannot be erected at Right Angles, to a given Plane, from a Point therein given ; which was to be demonstrated, 3 PROPOSITION XIV. THE OREM. perpendicular, are parallel to each otber. of the Planes D C, EF. I say, these Planes are parallel. For, if they be not, let them be produced till they meet each other, and let the Right Line GH be the common Section, in which take any Poiot K, and join A K, B K. Then, because A B is perpendicular to the Plane EF, it shall also be perpendicular to the Right Line B K, being in the Plane E F produced ; wherefore the Plane A B K is a Right Angle. And, for the same Reason, BAK is also a Right Angle. And 17. I. And so the two Angles ABK, BAK, of the Triangle A B K, are equal to two Right Angles, which is * impoffible : Therefore the Planes CD, EF, being produced, will not meet each other; and so are neceffarily parallel. Therefore, thefe Planes to which the same. Right Line is perpendicular, are parallel to each other; which was to be demonstrated. S! PROPOSITION XV. THEOREM. rallel to two Right Lines, touching one anotber, ET two Right Lines A B, BC, touching one ano ther, be parallel to two Right Lines D E, EF, touching one another, but not in the fame Plane with them. I say, the Planes passing through A B, BC, and DE, E F, being produced, will not meet each other. * 11 of this, For, let B G be drawn * from the Point B, perpendicular to the Plane passing through DE, EF, meeting the same in the Point G and through G let G H be 131. 1. drawn i parallel to ED, and KG parallel to EF; then, because B G is perpendicular to the Plane palDef.3. fing through D E, EF, it shall also make * Rignt An. gles with all Right Lines that touch it, and are in the Tame Plane. But GH and G K, which are both in the same Plane, touch it; therefore each of the Angles BGH, BGK, is a Right Angle. And fince B A is parallel to GH, the Angles GBA, BGH, are * equal to two Right Angles: But BGH is a Right Angle; wherefore G B A shall also be a Right Angle; and so BG is perpendicular to B'A. For the same Reason, GB is also perpendicular to BC; therefore fince a Right Line B G stands at Right An gles to two Right Lines B A, B Ç, mutually cutting 14 of ibis. each other; BG shall also be I at Right Angles to the Plane drawn thro' B A B C. But it is perpendicular to the Plane drawn through DE, EF; therefore BG 29. I. |