If two parallel planes be cut by a third plane, they have the same incli nation 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 straight lines EF and GH be the common section 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 also KN at right angles to EF in the plane AB : and through the straight lines KM, KN, let a plane be made to pass, cutting the plane CD in the line LO.. And because EF and GH are the common sections of the plane EH with the two parallel planes AB and CD, EF is parallel to GH (14. 2. Sup.). But EF is at right angles to the plane that passes through KN and KM (4. 2. Sup.), because it is at right angles to the lines KM and KN : therefore GH is also at right angles to the same plane (7. 2. Sup.), and it is therefore at right angles to the lines LM, LO which it meets in that plane. Therefore, since LM and LO are at right angles to LG, the common section of the two planes CD and EH, the angle OLM is the inclination of the plane CD to the plane EH (4. def. 2. Sup.). For the same reason the angle MKN is the inclination of the plane AB to the plane EH. But because KN and LO are parallel, being the common sections of the parallel planes AB and CD with a third plane, the interior angle NKM is equal to the exterior angle OLM (29. 1.); that is, the inclination of the plane AB to the plane EH, is equal to the inclination of the plane CD to the same plane EH. Therefore, &c. Q. E. D. If two straight lines be cut by parallel planes, they must be cut in the same ratio. Let the straight lines AB, CD be cut by the parallel planes GH, KL, MN, in the points A, E, B; C, F, D: As AĒ is to EB, so is CF to FD. 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 H plane EBDX, the common sections EX, BD, are parallel (14. 2. Sup.). А. For the same reason, because the two parallel planes GH, KL are cut G by the plane AXFC, the common sections AC, XF are parallel : And because EX is parallel to BD, a side of the triangle ABD, as AE to EB, so is (2. 6.) AX to XD. Again, be K cause XF is parallel AC, a side of the triangle ADC, as AX to XD, so is CF to FD : And it was proved that AX is to XD, as AE to EB : Therefore IN (11.5.), as AE to EB, so is CF to FD. Wherefore, if two straight lines, &c. BY Q. E. D. M E If a straight line be at right angles to a plane, every plane which passes through that line is at right angles to the first-mentioned plane. Let the straight line AB be at right angles to a plane CK; every plane which passes through AB is at right angles to the plane CK. Let any plane DE pass through AB, and let CE be the common section of the planes DE, CK ; take any point F in CE, from which draw FG in the plane DE at right angles to CE : And because AB is perpendicular to the plane CK, therefore D A it is also perpendicular to every A н straight line meeting it in that plane (1. def. 2. Sup.); and consequently it is perpendicular to CE: K Wherefore ABF is a right angle ; but GFB is likewise a right angle ; therefore AB is parallel (28. 1.) to FG. And AB is at right angles to the plane CK: therefore FG is F B E also at right angles to the same plane (7. 2. Sup.). But one plane is at right angles to another plane when the straight lines drawn in one of the planes, at right angles to their common section, are also at right angles to the other plane (def. 2. 2.); and any straight line FG in the plane DE, which is at right angles to CE, the common section of the planes, has been proved to be perpendicular to the other plane CK ; therefore the plane DE is at right angles to the plane CK. In like manner, it may be proved that all the planes which pass through AB are at right angles to the plane CK. Therefore, if a straight line &c. Q. E. D. PROP. XVIII. THEOR. If two planes cutting one another be each of them perpendicular to a third plane, their common section is perpendicular to the same plane. Let the two planes AB, BC be each of them perpendicular to a third plane, and BD be the common section of the first two; BD is perpendicular to the plane ADC. From D in the plane ADC, draw DE perpendicular to AD, and DE to DC. Because DE is perpendicular to AD, the common section of the planes AB and ADC; and because the plane AB is at right angles to ADC, DE is at B right angles to the plane AB (def. 2. 2. Sup.), and therefore also to the straight line BD in that plane (def. 1. 2. Sup.). For the same reason, DF is at right angles to DB. Since BD is therefore at right angles to both the lines DE and DF, it is at' right angles to the plane in which DE and DF are, that is, to the plane ADC (4. 2. Sup.). Wherefore, &c. Q. E. D. A F E C PROP. XIX. THEOR. Two straight lines not in the same plane being given in position, to draw a straight line perpendicular to them both. Let AB and CD be the given lines, which are not in the same plane ; it is required to draw a straight line which shall be perpendicular both to AB and CD. In AB take any point E, and through E draw EF parallel to CD, and let EG be drawn perpendicular to the plane which passes through EB, EF (10.2. Sup.). Through AB and EG let a plane pass, viz. GK, PROP. XVI. THEOR. If two straight lines be cut by parallel planes, they must be cut in the same ratio, 1 Let the straight lines AB, CD be cut by the parallel planes GH, KL, MN, in the points A, E, B; C, F, D: As AĒ is to ER, so is CF to FD. Join AC, BD, AD, and let AD meet the plane KL in the point X; H G EH so is (2. 6.) AX to XD. Again, be x K cause XF is parallel AC, a side of the triangle ADC, as AX to XD, so is CF to FD : And it was proved that AX is to XD, as AE to EB : Therefore N (11.5.), as AE to EB, so is CF to FD. Wherefore, if two straight lines, &c. BY Q. E. D. M PROP. XVII. THEOR. If a straight line be at right angles to a plane, every plane which passes through that line is at right angles to the first mentioned plane. Let the straight line AB be at right angles to a plane CK; every plane which passes through AB is at right angles to the plane CK. Let any plane DE pass through AB, and let CE be the common sèction of the planes DE, CK ; take any point F in CE, from which draw FG in the plane DE at right angles to CE: And because AB is perpendicular to the plane CK, therefore D G it is also perpendicular to every H straight line meeting it in that plane (1. def. 2. Sup.); and consequently it is perpendicular to CE : K Wherefore ABF is a right angle; but GFB is likewise a right angle ; therefore AB is parallel (28. 1.) to FG. And AB And AB is at right angles to the plane CK : therefore FG is с F B A PROP. XVIII. THEOR. If two planes cutting one another be each of them perpendicular to a third plane, their common section is perpendicular to the same plane. Let the two planes AB, BC be each of them perpendicular to a third plane, and BD be the common section of the first two ; BD is perpendicular to the plane ADC. From D in the plane ADC, draw DE perpendicular to AD, and DF to DC. Because DE is perpendicular to AD, the common section of the planes AB and ADC; and because the plane AB is at right angles to ADC, DE is at B right angles to the plane AB (def. 2. 2. Sup.), and therefore also to the straight line BD in that plane (def. 1. 2. Sup.). For the game reason, DF is at right angles to DB. Since BD is therefore at right angles to both the lines DE and DF, it is at right angles to the plane in which DE and DF are, that is, to the plane ADC (4. 2. Sup.). Wherefore, &c. Q. E. D. A F С PROP. XIX. THEOR. Two straight lines not in the same plane being given in position, to draw a straight line perpendicular to them both. Let AB and CD be the given lines, which are not in the same plane; it is required to draw a straight line which shall be perpendicular both to AB and CD. In AB take any point E, and through E draw EF parallel to CD, and let EG be drawn perpendicular to the plane which passes through EB, EF (10. 2. Sup.). Through AB and EG let a plane pass, viz. GK, |