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Book XI. Let AB, CD be two parallel straight lines, and let one

of them AB be at right angles to a plane ; the other CD is at right angles to the same plane.

Let AB, CD meet the plane in the points B, D, and join 57. 11. BD: Therefore& AB, CD, BD are in one plane. In the

plane to which AB is at right angles, draw DE at right angles to BD, and make DE equal to AB, and join BE, AE, AD. And because AB is perpendicular to the plane,

it is perpendicular to, every straight line which meets it, 3 Def. 11. and is in that planea : Therefore each of the angles ABD,

ABE is a right angle: And because the straight line BD

meets the parallel straight lines AB, CD, the angles ABD, 29.1. CDB are together equalb to two right angles : And ABD

is a right angle; therefore also CDB is a right angle, and CD perpendicular to BD: And because AB is equal to DE, and BD common, the two AB, BD are equal to the two ED, DB, and the angle ABD is

А

C equal to the angle EDB, because each

of them is a right angle; therefore the
• 4. 1. base AD is equal to the base BE :

Again, because AB is equal to DE,
and BE to AD; the two AB, BE, are
equal to the two ED, DA; and the Bl
base AE is common to the triangles

ABE, EDA; wherefore the angle
18. 1. ABE is equal to the angle EDA:

And ABE is a right angle; and there.
fore EDA is a right angle, and ED

E perpendicular to DA: But it is also perpendicular to BD; 4. 11. therefore ED is perpendiculare to the plane which passes f3 Def. 11. through BD, DA, and shallf make right angles with every

straight line meeting it in that plane : But DC is in the plane passing through BD, DA, because all three are in the plane in which are the parallels AB, CD: Wherefore ED is at right angles to DC; and therefore CD is at right angles to DE: But CD is also at right angles to DB; CD then is at right angles to the two straight lines DE, DB in the point of their intersection D: And therefore is at right anglese to the plane passing through DE, DB, which is the same plane to which AB is at right angles. Therefore, if two straight lines, &c. Q.E.D.

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Book XI.

PROP. IX. THEOR.

Two straight lines which are each of them parallel to the same straight line, and not in the same plane with it, are parallel to one another.

Let AB, CD, be each of them parallel to EF, and not in the same plane with it; AB shall be parallel to CD.

In EF take any point G, from which draw, in the plane passing through ÉF, AB, the straight line GH at right angles to EF; and in the plane passing through EF, CD, draw GK at right angles to the same EF. And because EF is perpendicular both to GH A JI and GK, Ė F is perpendiculara to

-B the plane HGK passing through them: and EF is parallel to AB;

6 therefore AB is at right angles Eto the plane HGK. For the same reason, CD is likewise at right angles to the plane HGK. There-o K D fore AB, CD, are each of them at right angles to the plane HGK. But if two straight lines are at right angles to the same plane, they shall be parallelc to one another. There-- 6. 11. fore AB is parallel to CD. Wherefore, two straight lines, &c. Q. E. D.

a 4. 11.

F 18.11.

PROP. X. THEOR.

If two straight lines meeting one another be paralJel to two others that meet one another, and are not in the same plane with the first two; the first two and the other two shall contain equal angles.

Let the two straight lines AB, BC, which meet one an- • other, be parallel to the two straight lines DE, EF, that meet one another, and are not in the same plane with AB, BC. The angle ABC is equal to the angle DEF.

Take BA, BC, ED, EF all equal to one another; and

• 33. ).

Book XI. join AD,

CF, BE, AC, DF: Because BA is equal and pamorallel to ED, therefore AD is a both equal and parallel to

BE. For the same reason, CF is equal
and parallel to BE. Therefore AD and
CF are each of them equal and parallel

А.
to BE. But straight lines that are pa-

rallel to the same straight line, and not 69. 11. in the same plane with

it, are parallel b to one another. Therefore AD is parale l Ax. 1. lel to CF; and it is equal to it, and

AC, DF join them towards the same
parts; and thereforea AC is equal and
parallel to DF. And because AB, BC

D
are equal to DE, EF, and the base AC
* 8. 1. to the base DF; the angle ABC is equald to the angle

DEF. Therefore, if two straight lines, &c. Q. E. D.

PROP. XI. PROB. To 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.

In the plane draw any straight line BC, and from the • 12. 1. point A drawa AD perpendicular to BC. If then AD be

also perpendicular to the plane BH, the thing required isalready done; but if it be pot,

A b 11. 1. from the point D drawb, in

E
the plane BH, the straight line
DE, at right angles to BC; and
from the point X, draw AF per-G F

н
pendicularto DE; and through
* 31. 1. F drawc GH parallel to BČ:

And because BC is at right an

gles to ED and DA, BC is at d 4. 11. right anglesd to the plane pass

B D C ing through ED, DA; and GH is parallel to BC. But, if

two straight lines be parallel, one of which is at right 'an* 8. 11. gles to a plane, the other shall be at righte angles to the

same plane; wherefore GH is at right angles to the plane *3 Def. 11. through ED, DA, and is perpendicularf to every 'straighit

line meeting it in that plane. But AF, which is in the

plane through ED, AD, meets it: Therefore GH is per. Book XI, pendicular to AF; and consequently AF is 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.

PROP. XII. PROB.

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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

iB at right angles to the plane. From any point B above the plane

• 11. 11. drawa BC perpendicular to it; and from A drawb AD parallel to BC.

31. 1. Because, therefore, AD, CB are two parallel straight lines, and one of

А A C them BC is at right angles to the given plane, the other AD is also at right angles to itc: 8. 11. Therefore a straight 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 cannot 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 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

Book XI. pass through BA, AC; the common section of this with

, the given plane is a straight a line passing through A: Let 23. 11.

- 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 an-
gles with every straight line
meeting it in that plane. But
DAE, which is in that plane,
meets CA; therefore CAE is a
right angle. For the same reason
BAE is a right angle. Where-
fore the angle CÃE is equal to D
the angle BAE; and they are in one plane, which is im-
possible. Also, from a point above a plane, there can be

but one perpendicular to that plane: for, if there could be 5 6. 11. two, they would be parallel b to one another, which is ab

surd. Therefore, from the same point, &c. Q. E. D.

PROP. XIV. THEOR. PLANES to which the same straight line is perpendicular, 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 straight line GH, in which take any point K, and join AK, BK: Then, because

AB is perpendicular to the plane "3 Def. 11. EF, it is perpendicular a to the

straight line BK which is in that
plane. Therefore ABK is a right
angle. For the same reason BAK
is a right angle; wherefore the two

B
angles ABK, BAK of the triangle

ABK are equal to two right angles,
17. 1. which is impossible b: Therefore

the planes CD, EF, though pro-
duced, do not meet one another;

· VD 8 Def. 11. that is, they are parallel . Therefore planes, &c. Q.E.D.

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