Book XI. PROP. XV. THEOR. IF two straight lines meeting one another, be pa- See Note. rallel to two straight lines which meet one another, but are not in the same plane with the first two; the plane which passes through these is parallel to the plane passing through the others. Let AB, BC, two straight lines meeting one another, be parallel to DE, EF, that meet one another, but are not in the same plane with AB, BC: the planes through AB, BC, and DE, EF shall not meet, though produced. From the point B draw BG perpendicular to the plane a 11. 11. which passes through DE, EF, and let it meet that plane in G; and through G draw GH parallel to ED, and GK pa-b 31. 1. rallel to EF: and because BG is perpendicular to the plane through DE, EF, it shall, make right angles with every straight line meeting it in that planes. But the straight lines GH, GK in that plane meet it: therefore each of the angles BGH, BGK is a right angle: and because BA is parallel to GH (for each of them is parallel to E D DE, and they are not both in the same plane with it) the angles 1 Book XI. PROP. XVI. THEOR. See Note. IF two parallel planes be cut by another plane, their common sections with it are parallels. Let the parallel planes, AB, CD be cut by the plane EFHG, and let their common sections with it be EF, GH: EF is parallel to GH. ཡ K F H B D For, if it be not, EF, GH shall meet, if produced, either on the side of FH, or EG: first, let them be produced on the side of FH, and meet in the point K; therefore, since 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 same reason K is also in the plane CD: wherefore the planes AB, CD produced meet one another; but they do not meet, since they are parallel by the hypothesis: therefore the straight lines EF, GH do not meet when produced on the side of FH; in the same manner it may be proved, that EF, GH do not meet when produced on the side of EG: but straight lines which are in the same 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. hypothesis: A C E G PRÓP. XVII. THEOR. IF two straight lines be cut by parallel planes, they shall 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 AE 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 plane EBDX, the common sections C Ha 16. 11. A EX, BD, are parallela. For the same reason, because the two Book XI. parallel planes GH, KL are cut by the plane AXFC, the common sections AC, XF are paral- G lel: and because EX is parallel to BD, a side of the triangle ABD, as AE to EB, so is b AX to XD. Again, because XF is parallel to AC, a side of the triangle ADC, as AX to XD, so is CF to FD: and it was K E F b 2.6. PROP. XVIII. THEOR. IF a straight line be at right angles to a planc, evèry plane which passes through it shall be at right angles to that plane. Let the straight line AB be at right angles to a plane CK; every plane which passes through AB shall be 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 it is also perpendicular to every straight line in that plane meeting it ; and consequently it is perpendicular to CE: wherefore ABF is a right angle; but GFB is likewise a right angle: there D G A H K a3. def. 11. BE fore AB is parallel b to FG. And AB is at right angles to the b 28. 1. plane CK; therefore FG is also at right angles to the same plane. But one plane is at right angles to another plane when c 8, 17. the straight lines drawn in one of the planes, at right angles to their common section, are also at right angles to the other Book XI. plane d: and any straight line FG in the plane DE, which is at right angles to CE the common section of the planes, has 4. def. 11. 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. XIX. THEOR. IF two planes cutting one another be each of them perpendicular to a third plane; their common section shall be perpendicular to the same plane. Let the two planes AB, BC be each of them perpendicular to a third plane, and let BD be the common section 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 straight line DE at right angles to AD the common section of the plane AB with the third plane; and in the plane BC draw DF at right angles to CD the common section of the plane BC with the third plane And because the plane AB is perpendicular to the third plane, and DE is drawn in the plane AB at right angles to AD their common section, DE is perpendicular to the third a 4. def. 11. plane a. In the same manner, it may be proved that DF is perpendicular to the third plane. Wherefore, from the point D two straight lines stand at right angles to the third plane, upon the same side of it, which is impossible b: therefore, from the point D there cannot be any straight line at right angles to the third plane, A except BD the common section of the 13.11. D planes AB, BC. BD therefore is perpendicular to the third plane. Wherefore, if two planes, &c. Q. E. D. Book XI. PROP. XX. THEOR. IF a solid angle be contained by three plane angles, See Note. any two of them are greater than the third. Let the solid 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 be 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 a to the angle DAB; and make AE equal to AD, and through E draw BEC cutting AB, AC in a 23. 1. 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 base DB is equal b to the base BE. And because BD, DC are greater than CB, and one of them BD has been proved equal to BE a part of CB, therefore the other DC is greater than the remaining part EC. And B because DA is equal to AE, and AC A b 4. 1. c 20. 1. E C common, but the base DC greater than the base EC: therefore the angle DAC is greater than the angle EAC; and, by the construction, the angle DAB is equal to the angle BAE; where- d 25. 1. fore the angles DAB, DAC are together greater than BAE, EAC, that is, than the angle BAC. But BAČ is not less than either of the angles DAB, ĎAC: therefore BAC, with either of them, is greater than the other. Wherefore, if a solid angle, &c. Q. E. D. PROP. XXI. THEOR. EVERY solid angle is contained by plane angles which together are less than four right angles. First, let the solid angle at A be contained by three plane angles BAC, CAD, DAB. These three together are less than four right angles. |