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'3 Def. 11.
PROP. XV. THEOR. If two straight lines meeting one another, be pa- See N. 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 perpendiculara to the plane a 31. 11. which passes through DE, EF, and let it meet that plane in G; and through G draw GH parallel b to ED, and GK • 31. 1.. parallel 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 planec. But the? straight lines GH, GK in Bio that plane meet it: Therefore each of the angles BGH, BGK is a right angle : And because BA is parallel d to AL. GH (for each of them is parallel to DE, and they are not ; both in the same plane with it), the angles GBA, BGH are together equale to two right angles : And BGH is a e 29. In right angle; therefore also GBA is a right angle, and GB perpendicular to BA: For the same reason, GB is perpendicular to BC: Since therefore the straight line GB stands at right angles to the two straight lines BA, BC, that cut one another in B; GB is perpendicularf to the plane f 4. 11. through 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 same straight line is perpendicular, are parallel s to one another: Therefore the plane through AB, BC • 14. 11. is parallel to the plane through DE, EF. Wherefore if two straight lines, &c. Q. E. D.
d 9. 11.
PROP. XVI. THEOR. See N. 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.
For, if it is 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 AS 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.
PROP. 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 parallelplanes 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 EX, BD are parallela. For the same reason, be- Boox XI. cause the two parallel planes
a 16. 11. GH, KL are cut by the plane I AXFC, the common sections AC, XF are parallel : And because EX is parallel to BD, a side of the triangle ABD;
12.6. as AE to EB, so is LAX to .. XD. Again, because XF is parallel to AC, a side of the KL 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, as MAE to EB, so is CF to FD. Wherefore, if two straight lines, &c. Q. E. D.
c 11. 5.
PROP. XVIII. THEOR.
If a straight line be at right angles to a plane, every 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 an.. gles 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 D G
H 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 ita : And consequently it is is perpendicular to CE: Wherefore ABF is a right angle; but GFB is likewise a right angle; therefore AB is parallel b to FG. And AB is at right angles to the plane CK; b 28. 1. therefore FG is also at right angles to the same plane C. c 3. 11. But one plane is at right angles to another plane when the straight lines drawn in one of the planes, at right angles to
.3 Def. 11.
4 Def. 1
Book XI. their 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 CE the conimon section of the planes, lias 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 sec. tion 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.
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 B ETH 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 perpen* 4 Def. 11. dicular to the third planea. In the
same manner, it may be proved, that
the thirrl plane, upon the same side of it,
the point D there cannot be any straight line at right angles
· PROP. XX. THEOR.
If a solid angle be contained by three plane angles, See N. 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.
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 a a 23. 1. to the angle DAB; and make A E 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 base DB is equal b to the base BE. And be- B
E Co4. 1. cause BD, DC are greater than CB, and one of them BD « 20. 1. 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 base EC; therefore the angle DAC is greater d than the angle EAC; and, by the construction, 25. 1. 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, 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