Book XI. fore AD is a both equal and parallel to

B
BE. For the famé reason, CF is equal
a. 33. I.

and parallel to BE. Therefore AD and
CF are each of them equal and parallel A

C to BE. But straight lines that are paral

lel to the same straight line, and not in b. 9. II. the same plane with it, are parallel to

one another. Therefore AD is parallel C. 1. Ax. I. to CF; and it is equal < to it, and AC,

E
DF join them towards the same parts; D

F and therefore a AC is equal and parallel

to DF. and because AB, BC are equal to DE, EF, and the base d. 8. I.

AC to the base DF; the angle ABC is equal d to the angle DEF.
Therefore if two straight lines, &c. Q. E. D.

To a

10 draw a straight line perpendicular to a plane,

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.

b. 11. I.

In the plane draw any straight line BC, and from the point A draw a AD perpendicular to BC. If then AD be also perpendiçular to the plane BH, the thing required is already done; but if it be not, from the point D draw

A
in the plane BH, the straight line
DE at right angles to BC; and from E
the point A draw AF perpendicular G

H C. 31. I. to DE; and thro' F draw GH pa

rallel to BC. and because BC is at

right angles to ED and DA, BC is d. 4. II. at right angles d to the plane pafsing thro' ED, DA. And GH is

B D parallel to BC; but if two straight lines be parallel, one of which is at right angles to a plane, the other shall be at right e angles to

the same plane; wherefore GH is at right angles to the plane thro' €. 3. Def. 11. ED, DA, and is perpendicular f to every straight line meeting it in

that planę. But AF, which is in the plane thro' ED, DA meets it.

e. 8. II.

therefore GH is perpendicular to AF, and consequently AF is Book XI. perpendicular to GH. 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.

10 ere& a straight line at right angles to a given

plane, from a point given in the plane.

1

Let A be the point given in the plane; it is required to erect a straight line from the point A at right an.

D B gles to the plane.

From any point B above the plane draw a BC perpendicular to it; and from A draw 6 AD parallel to BC. because therefore AD, CB are two parallel straight A c lines, and one of them BC is at right angles to the given plane, the other AD is also at right angles to it c. therefore a straight line has been erect- C. 8. II. ed at right angles to a given plane from a point given in it. Which was to be done.

2. II. II.

b. 31. 1.

FR

PROM the same point in a given plane there cannot

be two straight lines at right angles to the plane, upon

the same side of it. and there can be but one per. pendicular to a plane from a point above the plane.

For, if it be possible, let the two straight lines AB, AC 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 thro' BA, AC; the common section of this with the given plane is a straight a line 2. 3. BIa

Book XI. paling through A. let DAE be their common section. therefore

the straight lines AB, AC, DAE are in one plane. and because
CA is at right angles to the given plane, it shall make right angles
with every straight line meeting it in
that plane. but DAE which is in

B C
that plane meets CA; therefore CAE
is a right angle. for the same reason
BAE is a right angle. wherefore the
angle CAE is equal to the angle
BAE; and they are in one plane,

D А A E which is impossible. Also, from a point above a plane there can

be but one perpendicular to that plane; for if there could be two, b. 6. 11. they would be parallel to one another, which is absurd. There

fore from the same point, &c. Q. E. D.

PLANES

LANES to which the same fraight line is perpen.

dicular, 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 meet; their common section shall be a

G
straight line GH, in which take any
point K, and join AK, BK. then be-

cause AB is perpendicular to the plane
2. 3. Def. 11. EF, it is perpendicular a to the straight C

H line BK which is in that plane. there

F fore ABK is a right angle. for the same

A
reason, BAK is a right angle; where-
fore the two angles ABK, BAK of the

triangle ABK are equal to two right b. 19. s. angles, which is impöllible b. therefore

the planes CD, EF though produced do c. 8.Def.tr. not meet one another; that is, they are parallel c. Therefore

planes, &c. Q. E. D.

B

E

Book XI.

PROP. XV.

THEOR.

IF

two straight lines meeting one another, be parallel See N.

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 pas. sing 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 fhall not meet though produced.

From the point B draw BG perpendicular to the plane which a. 11. it. passes through DE, EF, and let it meet that plane in G; and through G draw. GH parallel o to ED, and GK parallel to EF. b. 31. t. and because BG is perpendicular to the plane through DE, EF, it shall make right angles with every straight line meeting it in that plane c. but the straight B

G

F c.3.Def.11. lines GH, GK in that plane

C

К. meet it. therefore each of the angles BGH, BGK is a right angle. and because BA is paral- A lel d to GH (for each of them is

H

d. 9.11. parallel to DE, and they are not both in the same plane with it) the angles GBA, BGH are together equal o to two right angles. and BGH is a right angle, e. 29. 1. therefore also GBA is a right angle, and GB perpendicular to BA. for the fame 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 perpendicular f f. 4. 18. to the plane 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 fame straight line is perpendicular, are parallel 8 to one another. 8. 14. 11. therefore the plane thro' AB, BC is parallel to the plane thro' DE, EF. Wherefore if two straight lines, &c. Q. E. D.

Book XI.

2

PROP. XVI." THEOR.

See N.

[F two parallel planes be cut by another plane, their

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
fide of FH, or EG. first, let them be produced on the side 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;

К.
therefore K is in the plane AB.
for the same reason K is also in
the plane CD. wherefore the

F E

H planes AB, CD produced meet

B

D
one another; but they do not
meet, sinçe they are parallel by
the Hypothesis. therefore the A
straight lines EF, GH do not E
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 ftraight lines
which are in the same plane and do not meet, though produced
either way, are parallel. therefore EF is parallel to GH. Where-
fore if two parallel planes, &c. Q. E. D.

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, 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 cut by the plane EBDX, the common sections EX, BD are paral