IF PROP. XV. THEOR. Book XI. two straight lines meeting one another, be parallel to See N. two straight 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 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 tho' produced. From the point B draw BG perpendicular to the plane which a. 11. 11. paffes through DE, EF, and let it meet that plane in G; and through G draw GH parallel b to ED, and GK parallel to EF. and b. 31. 1. because BG is perpendicular to the plane through DE, EF, it shall both in the fame plane with it) the angles GBA, BGH are together equal to two right angles. and BGH is a right angle, therefore e. 29. 1. also GBA is a right angle, and GB perpendicular to BA. for the fame reason, GB is perpendicular to BC. fince therefore the straight line GB ftands at right angles to the two straight lines BA, BC, that cut one another in B; GB is perpendicular to the plane through f. 4. 11: BA, BC. and 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 straight line is perpendicular, are parallel & to one another. therefore the plane through g. 14. 11. AB, BC is parallel to the plane through DE, EF. Wherefore if two straight lines, &c. Q. E. D. PROP. Book XI. Sec N. PROP. XVI. THEOR. [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 A E F B H meet when 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 straight 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. PROP. XVII. THEOR. two straight lines be cut by parallel planes, they fhall be cut in the fame 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 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 a. 16. 1. cut by the plane EBDX, the common fections EX, BD are parallel®. for for the fame reafon, because the two parallel planes GH, KL are Book XI. cut by the plane AXFC, the IF PROP. XVIII. THEOR. F a straight line be at right angles to a plane, every plane which paffes through it fhall be at right angles to that plane. Let the straight line AB be at right angles to the plane CK. every plane which paffes through AB fhall be at right angles to the plane CK. Let any plane DE pass through AB, and let CE be the common fection of the planes DE, CK; take any point F in CE, from which D G A H K T a. 3. Def.11. draw FG in the plane DE at C F BE common Book XI. common section, are also at right angles to the other planed; and any straight line FG in the plane DE, which is at right angles to 4.4.Def.11.CE the common fection of the planes, has been proved to be per pendicular 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. XIX. THEOR. 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 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 becaufe the plane AB is perpendicular to the third plane, and DE is drawn in the plane AB at right angles to AD their common fection, DE is perpendicular to the third a.4.Def.11. plane. in the fame manner, it may be pro ved that DF is perpendicular to the third b. 13. 11. is impoffible b. therefore from the point D E D C there cannot be any straight line at right angles to the third plane, except BD the common fection 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. 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. 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 less than either of the other two, and is greater than one of them DAB; and at the point A in the straight line AB, make in the plane which passes through BA, AC, the angle BAE equal to the angle DAB; and make AE equal to a. 23. 1. AD, and through E draw BEC cutting a D AB, AC in the points B, C, and join DB, DC. and becaufe DA is equal to C B E C b. 4. 1. 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 b to the bafe BE. and because BD, DC are greater than CB, and one of them BD has been proved equal c. 20. I. to BE a part of CB, therefore the other DC is greater than the remaining part EC. and becaufe DA is equal to AE, and AC common, but the base DC greater than the bafe EC; therefore the angle DAC is greater than the angle EAC; and, by the conftruction, the d. 15. 1. angle DAB is equal to the angle BAE; wherefore the angles DAB, DAC are together greater than the angle BAC. but BAC is not lefs than either of the angles DAB, DAC, therefore BAC with either of them is greater than the other. Wherefore if a folid angle, &c. QE. D. PROP. XXI. THEOR. EVERY folid angle is contained by plane angles which' together are less than four right angles, First, Let the folid angle at A be contained by three plane angles BAC, CAD, DAB. these three together are lefs than four right angles. Take |