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Book XI. fore AD is a both equal and parallel to BE.
For the same reason CF is equal and paa 33. 1. rallel to BE. Therefore AD and CF are
each of them equal and parallel to BE.
same straight line, and not in the same b 9. 11. plane with it, are parallel b to one ano
ther. Therefore AD is parallel to CF; cl. Ax.l. and it is equal c to it, and AC, DF join
them towards the same parts; and there-
EF, and the base AC to the base DF; the d 8. 1.
angle ABC is equal 4 to the angle DEF.
PROP. XI. PROB.
TO draw a straight line perpendicular to a planc, from a given point above it.
Let A be the given point above the plane BH; it is required to draw from the point A a straight line perpendicular to the plane BH.
In the plane draw any straight line BC, and from the point A a 12. 1. draw a AD perpendicular to BC. If then AD be also perpendi
cular to the plane BH, the thing required is already done; but if b 11. 1. it be not, from the point D drawb, in the plane BH, the straight
line DE at right angles to BC: and from the point A draw AF c 31. 1. perpendicular to DE ; and through F drawcGH parallel to BC: and because BC is at right angles
H parallel, one of which is at right
angles to a plane, the other shall e 8. 11. be at right e angles to the saine
plane; wherefore GH is at right
с f3. def. DA, and is perpendicular f to 11.
every straight line meeting it in that plane. But AF, which is in the plane through ED, DA, meets it: therefore GH is perpendicular to AF; and consequently AF is perpendicular to GH; Book XI. and AF is perpendicular to DE: therefore AF is perpendicular to each of the straight lines GH, DE. But if a straight line stands at right angles to each of two straight lines in the point of their intersection, it shall also be at right angles to the plane passing through them. But the plane passing through ED, GH is the plane BH; therefore AF is perpendicular to the plane BH; therefore, from the given point A, above the plane BH, the straight line AF is drawn perpendicular to that plane. Which was to be done.
PROP. XII. PROB.
TO erect a straight line at right angles to a given plane, from a point given in the plane.
Let A be the point given in the plane ; it is required to erect a straight line from the point A at right
D angles to the plane.
B From any point B above the plane draw a
a 11. 11. BC perpendicular to it; and from A drawb
b 31. 1. AD parallel to BC. Because, therefore, AD, CB are two parallel straight lines, and one of them BC is at right angles to
А с the given plane, the other AD is also at right angles to its. Therefore a straight
c 8. 11. line has been erected at right angles to a given plane from a point given in it. Which was to be done.
PROP. XIII. THEOR.
FROM the same point in a given plane, there can. not be two straight lines at right angles to the plane, upon the same side of it; and there can be but one perpendicular to a plane from a point above the plane.
For, if it be possible, let the two straight lines AC, AB be at right angles to a given plane from the same point A in the plane, and upon the same side of it; and let a plane pass through BA,
Book XI. AC; the common section ofthis with the given plane is a straight
line passing through A: let DAE be their common section: therea 3. 11. fore the straight lines AB, AC, DAE are in one plane: and be
cause CA is at right angles to the given plane, it shall make
E b 6.11. be parallel b to one another, which
is absurd. Therefore, from the same point, &c. Q. E. D.
PROP. XIV. THEOR.
PLANES to which the same straight line is pendicular, are parallel to one another.
Let the straight line AB be perpendicular to each of the planes CD, EF; these planes are parallel to one another.
If not, they shall meet one another when produced ; let them mect; their common section shall be a straight line GH, in which take any
G point K, and join AK, BK : then, be
cause AB is perpendicular to the plane a 3. def. EF, it is perpendicular a to the straight 11. line BK which is in that plane. There
C fore ABK is a right angle. For the
н same reason, BAK is a right angle ;
F wherefore the two angles ABK, BAK
B of the triangle ABK are equal to two b 17. 1. right angles, which is impossible b: therefore the planes CD, EF, though
E produced, do not meet one another; . 8. def. that is, they are parallel c. Therefore,
planes, &c. Q. E. D.
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 perpendicular a 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 pa-b 31. 1. rallel to EF : and because BG is perpendicular to the plane through DE, EF, it shall
E make right angles with every straight line meeting it in that
G F plane c. But the straight lines B
c 3. def. GH, GK in that plane meet
K 11. it: therefore each of the angles BGH, BGK is a right an. gle: and because BA is pa
D rallel d to GH (for each of A
d 9.11. them is parallel to DE, and
Я they are not both in the same plane with it) the angles GBA, BGH are together cquale to two right angles: and BGH is a e 29. 1. 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 through BA, BC ; and it is f 4. 11. 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 8 to one another : therefore the plane through AB, 8 14. 11.) BC is parallel to the plane through DE, EF. Wherefore, if two straight lines, &c. Q. E. D.
PROP. XVI. THEOR.
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
D meet one another; but they do not meet since they are
C parallel by the hypothesis :
A therefore the straight lines
G 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 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