PROP. XV. THEOR. Book XI. two straight lines meeting one another, be parallel to See N. two ftraight lines which meet one another, but are not in the fame plane with the first two; the plane which paffes through these is parallel to the plane paffing through the others. Let AB, BC, two ftraight lines meeting one another, be pa rallel 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. a From the point B draw BG perpendicular to the plane a II. II. which paffes through DE, EF, and let it meet that plane in G; and through G draw GH parallel b to ED, and GK pa- b 31. 1. rallel to EF: And because BG is perpendicular to the plane through DE, EF, it fhall E make right angles with every ftraight line meeting it in that B. plane But the ftraight lines GH, GK in that plane meet it: Therefore each of the angles BGH, BGK is a right angle: And because BA is parallel d to GH (for each of them is parallel to DE, and с с they are not both in the fame plane with it) the angles GBA, BGH are together equal to two right angles: And BGH is a e 29. I. right angle; therefore alfo GBA is a right angle, and GB perpendicular to BA: For the fame reafon, GB is perpendicular to BC: Since therefore the ftraight 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 f 4. II. it 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 to one another: Therefore the plane through AB, g 14. 11. BC is parallel to the plane through DE, EF. Wherefore, if two ftraight lines, &c. QE. 1). PROP Book XI. See N. IF F two parallel planes be cut by another plane, their common fections with it are parallels. 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. K For, if it is not, EF, GH fhall meet, if produced, either on the fide of FH, or EG: First, let them be produced on the fide of FH, and meet in the point K: Therefore, fince EFK is in the plane AB, every point in EFK is in that plane; and K is a point in EFK; therefore K is in the plane AB: For the fame reafon K is alfo in the plane CD: Wherefore the planes AB, CD produced meet one another; but they do not meet, fince they are parallel by the hypothefis : Therefore the ftraight lines EF, GH do not meet when A E F H B D G produced on the fide of FH: In the fame manner it may be proved, that EF, GH do not meet when produced on the fide of EG: But ftraight lines which are in the fame plane and do not meet, though produced either way, are parallel: Therefore EF is parallel to GH. Wherefore, if two parallel planes, &c. Q. E. D. IF two straight 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. 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 parallel a. For the fame reason, because the two Book XI. parallel planes GH, KL are cut by the plane AXFC, the common fections AC, XF are K M IE E a ftraight line be at right angles to a plane, every plane which paffes through it fhall be at right angles to that plane. Let the ftraight line AB be at right angles to a plane CK; every plane which paffes through AB fhall be at right angles to the plane CK. D Let any plane DE pafs through AB, and let CE be the common fection 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 it is alfo perpendicular to every ftraight line in that plane meeting it a ; And confequently it is perpen. dicular to CE: Wherefore ABF is a right angle; but GFB is likewife a right angle; therefore AB is parallel to FG. G A H K a 3. def. 1 C F BE And AB is at right angles to b 28. I, the plane CK; therefore FG is alfo at right angles to the fame plane c. But one plane is at right angles to another plane when c 8. 11. the ftraight lines drawn in one of the planes, at right angles d 4. def. II. Book XI. to their common fection, are alfo at right angles to the other planed; and any straight line FG in the plane DE, which is at right angles to CE the common fetion 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 pafs through AB are at right angles to the plane CK. Therefore, if a straight line, &c. Q. E. D. IF two planes cutting one another be each of them perpendicular to a third plane; their common fection fhall be perpendicular to the fame plane. Let the two planes AB, BC be each of them perpendicular to a third plane, and let BD be the common fection of the first two; BD is perpendicular to the third plane. B E F If it be not, from the point D draw, in the plane AB, the ftraight line DE at right angles to AD the common fection of the plane AB with the third plane; and in the plane BC draw DF at right angles to CD the common fection of the plane BC with the third plane. And because the plane AB is perpendicular to the third. plane, and D is drawn in the plane AB at right angles to AD their common. fection, DE is perpendicular to the third 24. def. 11. plane a. In the fame manner, it may be proved that DF is perpendicular to the third plane. Wherefore, from the point D two ftraight lines ftand at right angles to the third plane, upon the fame b 13. 11. fide of it, which is impoffible b: Therefore, from the point D there cannot be any ftraight line at right angles to the A third plane, except BD the common fec D C tion of the planes AB, BC. BD therefore is perpendicular to the third plane. Wherefore, if two planes, &c. Q. E. D. PROP PROP. XX. THEOR. Book XI. IF Fa folid angle be contained by three plane angles, any See N two of them are greater than the third. Let the folid angle at A be contained by the three plane angles BAC, CAD, DAB. Any two of them are greater than the third. D If the angles BAC, CAD, DAB be all equal, it is evident that any two of them are greater than the third. But if they are not, let BAC be that angle which is not lefs than either of the other two, and is greater than one of them DAB; and at the point A in the ftraight line AB, make, in the plane which paffes through BA, AC, the angle BAE equal to thea 23. 1. angle DAB; and make AE equal to AD, and through E draw BEC cutting AB, AC in the points B, C, and join DB, DC. And because DA is equal to AE, and AB is common, the two DA, AB are equal to the two EA, AB, and the angle DAB is equal to the angle EAB: Therefore the bafe DB is equal to the bafe BE. And becaufe BD, DC B are greater than CB, and one of them b 4. I. E C c 20. Í BD has been proved equal to BE a part of CB, therefore the other DC is greater than the remaining part EC. And because DA is equal to AE, and AC common, but the base DC greater than the bafe EC; therefore the angle DAC is greater than d. 25. 1. the angle EAC; and, by the conftruction, the angle DAB is equal to the angle BAE; wherefore the angles DAB, DAC are together greater than BAE, EAC, that is, than the angle BAC. But BAC is not less than either of the angles DAB, DAC; therefore BAC, with either of them, is greater than the other. Wherefore, if a folid angle, &c. Q.E. D. VERY folid angle is contained by plain angles which Firft, Let the folid angle at A be contained by three plane angles BAC, CAD, DAB. These three together are less than four right angles. 0 3 Take |