Supplement would be parallel b to one another, which is abfurd. Therefore, from the fame point, &c. Q. E. D. b.6.7. PROP. XII. THEOR. LANES to which the fame ftraight line is perpendicular, are parallel to one another. PL C G K H F A B Let the ftraight line AB be perpendicular to each of the planes CD, EF; these planes are parallel to one another. If not, they shall meet one another when produced; let them meet: their common section shall be a straight line GH, in which take any point K, and join AK, BK: Then, because AB is perpendicular to the plane EF, it is perpendicular a to the ftraight line BK which is in that plane. Therefore ABK is a right angle. For the fame reafon, BAK is a right angle; wherefore the two angles ABK, BAK of the triangle ABK are equal to two right angles, d 17. 1. which is impoffible b: Therefore the planes CD, EF, though produced, do a 1. def. 2. Sup. E D def. 2. not meet one another; that is, they are parallel c. Therefore planes, &c. Q.E. D. c "2 Sup⋅ PROP. XIII. THEOR. F two ftraight lines meeting one another, be parallel to two ftraight lines which meet one another, but are not in the fame plane with the first two; the plane which paffes through thefe is parallel to the plane paffing through the others. Let Let AB, BC, two ftraight lines meeting one another, be Book II. parallel to DE, EF that meet one another, but are not in the fame plane with AB, BC: The planes through AB, BC, and DE, EF fhall not meet, though produced. b 31. 1. From the point B draw BG perpendicular a to the plane a 10.2. Sup. which paffes through DE, EF, and let it meet that plane in G; and through G draw GH parallel to ED b, and GK parallel to EF: And becaufe BG is perpendicular to the plane through DE, EF, it shall make right angles with every ftraight line meeting it in that plane c, But the ftraight lines GH, B GK in that plane meet it: Therefore each of the angles BGH, BGK is a right angle: And because BA is paralleld to GH (for each of them is parallel to DE), the angles GBA, BGH are toge E G F c 1 def. z. Sup. K C d 8. 2. Sup, D H . ther equal e to two right angles: And BGH is a right angle; e 29. 1. therefore alfo GBA is a right angle, and GB perpendicular to BA: For the fame reafon, GB is perpendicular to BC: Since, therefore, the straight line GB ftands at right angles to the two ftraight lines BA, BC, that cut one another in B; GB is perpendicular f to the plane through BA, BC: And it f4. 2. Sup. is perpendicular to the plane through DE, EF; therefore BG is perpendicular to each of the planes through AB, BC, and DE, EF: But planes to which the fame ftraight line is perpendicular, are parallel g to one another: Therefore the plane g12. 2. Sup. through AB, BC is parallel to the plane through DE, EF. Wherefore, if two straight lines, &c. Q. E. D. COR. It follows from this demonftration, that if a ftraight. line meet two parallel planes, and be perpendicular to one of them, it must be perpendicular to the other also. Supplement PROP. XIV. THEOR. F two parallel planes be cut by another plane, their common fections with it are parallels. IF Let the parallel planes AB, CD be cut by the plane EFHG, and let their common fections with it be EF, GH; EF is parallel to GH. For the ftraight lines EF and GH are in the fame plane, viz. EFHG, which cuts the planes AB and CD; and they do not meet though produced; for the planes in which they are do not meet; therefore EF and 830. def. 1. GH are parallel a. Q E. D. A E F H D B F two parallel planes be cút by a third plane, they have the fame inclination to that plane. Let AB and CD be two parallel planes, and EH a third plane cutting them: The planes AB and CD are equally inclined to EH. Let the ftraight lines EF and GH be the common fectiou of the plane EH with the two planes AB and CD; and from K, any point in EF, draw in the plane EH the straight line KM at right angles to EF, and let it meet GH in L; draw alfo KN at right angles to EF in the plane AB: and through the ftraight lines KM, KN, let a plane be made to pafs, cutting the plane CD in the line LO. And becaufe EF and GH are the common fections of the plane EH with the two two parallel planes AB and CD, EF is parallel to GH a. Book II. But EF is at right angles to the plane that paffes through a14. 2. Sup. KN and KM b, be b 4. 2. Sup. LO are at right 4. def. 2. Sup. angles to LG, the common fection of the two planes CD and EH, the angle OLM is the inclination of the plane CD to the plane EH d. For the fame reason the angle MKN is the in-d clination of the plane AB to the plane EH. But because KN and LO are parallel, being the common fections of the parallel planes AB and CD with a third plane, the interior angle NKM is equal to the exterior angle OLM e: that is, e 29. 1. the inclination of the plane AB to the plane EH, is equal to the inclination of the plane CD to the fame plane EH. Therefore, &c. Q. E. D. Supplement PROP. XVI. THEOR. I F two ftraight lines be cut by parallel planes, they fhall be cut in the fame ratio. Let the ftraight lines AB, CD be cut by the parallel planes GH, KL, MN, in the points A, E, B; C, F, D: As AE is to EB, fo is CF to FD. H C A G Join AC, BD, AD, and let AD meet the plane KL in the point X; and join EX, XF : Because the two parallel planes KL, MN are cut by the plane EBDX, the common fections EX, BD, are 214. 2. Sup. parallel a. For the fame reafon, because the two parallel planes GH, KL are cut by the plane AXFC, the common fections AC, XF are parallel: And because EX is parallel to BD, a fide of the triangle ABD, as AE to EB, 2. 6. fo is b AX to XD. Again, because XF is parallel to AC, a fide of the triangle ADC, as AX to XD, fo is CF to FD: And it was pro F E KL X B D M c 11.5. ved that AX is to XD, as AE to EB: Therefore, as AE to EB, fo is CF to FD. Wherefore, if two straight lines, Q. E. D. &c. I' F a straight line be at right angles to a plane, every plane which paffes through it is alfo at right angles to that plane. 1 Let |
