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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 the others.
Let AB, BC, two straight lines meeting one another, be pa. rallel 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 11.11. which passes through DE, EF, and let it meet that plane in G; and through G draw GH parallel b to ED, and GK, paral-b 31.1, lel to EF: and because BG is perpendicular to the plane through DE, EF, it shall
E make right angles with every
F straight line meeting it in
G that planec. But the straight B
K lines GH, GK in that plane
c 3 def. H. meet it: therefore each of the angles BGH, BGK is a right angle: and because BA A
D is parallel to GH (for each
d 9. 14 of them is parallel to DE, and they are not both in the same plane with it) the angles GBA, BGH are together equal e to two right angles: and BGH is e 29.1, a right angle; therefore also GBA is a right angle, and GB perpendicular to BA: for the same reason, GB is perpendicu. lar 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 perpendiculars to the plane through BA, BC: and f 4. 11. 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 parallels to one another: therefore the plane through AB, BC is parallel to the plane through DE, EF. Wherefore, if two straight lines, &c. Q. E. D.
PROP. XVI. THEOR.
Sée Notes 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 planc EFHG, and let their common sections with it be EF, GH: EF is parallel to GH.
For, if it be not, EF, GH shall meet, if produced, either 911
D produced meet one another; but they do not meet, since they are ch parallel by
the hypothesis: A 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 meets though produced either way, are parallel : therefore EF is parallel to GH. Wlrerefore, 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 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 KI., MN are cut by the plane EBDX, the common sections
EX, BD, are parallel a, For the same reason, because the two Book XI,
H. 16. 11.
PROP. XVIII. THEOR.
IF a straight line be at right angles to a plane, eve: ry plane which passes through it shall be at right an. gles 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 com-
G А H
a 3. def. u. consequently it is perpendicular to CE: wherefore ABF is
F В. E right angle; but GFB is likewise a right angle: therefore AB is parallel to FG. And AB is at right angles to the b 28.2 plane CK; therefore FG is also at right angles to the same planec. But one plane is at right angles to another plane when c8 11. 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. planed: and any straight line FG in the plane DE, which is
at right angles to CE the common section of the planes, has d 4. def. 11. been proved to be perpendicular to the other plane CK; there
fore 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.
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 B plane AB is perpendicular to the third plane, and DE is drawn in the planc AB at vight angles to AD their common section, DE is perpendicular to the third
E F 24. def. 11. plane a. In the same manner, it may be
proved that DF is perpendicular to the
to the third plane, upon the same side of 5 13. 11.
it, which is impossible b: therefore, from
C line at right angles to the third plane, except BD the common section of the planes AB, BC. BD therefore is perpendicular to the third plane. Wherefore, if two planes, &c. Q. E. D.
IF a solid angle be contained by three plane angles, 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 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 a 23. 1. 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 D 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 equalb to the base BE.
b 4. 1.
c 20. 1:
E c therefore the other DC is greater than the remaining part EC. And because DA is equal to EA, and AC common, but the base DC greater than the base EC: therefore the angle DAC is greaterd than d 25. 1. the angle EAC; and, by the construction, 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 angleş 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 plane angles BAC, CAD, DAB. These three together are less than four right angles.