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BE to AD; AB, BE are equal to ED, DA; and, in the Book XI.
triangles ABE, EDA, the base AE is common: therefore
the angle ABE is equal to the angle EDA: But ABE is * 8. 1.
a right angle: therefore EDA is also a right angle, and
ED perpendicular to DA: But it is also perpendicular to
each of the two BD, DC: Wherefore ED is at right angles
to each of the three straight lines BD, DA, DC in the point
in which they meet: Therefore these three straight lines
are all in the same planea: But AB is in the plane in which d 5. 11.
are BD, D1, because any three straight lines which meet
one another are in one planee: Therefore AB, BD, DC, 2. 11.
are in one plane: And each of the angles ABD, BDC is a
right angle; therefore AB is parallelf to CD. Wherefore,
if two straight lines, &c. Q. E.D.

f28. 1.

PROP. VII. THEOR.

1

If two straight lines be parallel, the straight line See N.
drawn from any point in the one to any point in
the other, is in the same plane with the parallels.

Let AB, CD be parallel straight lines, and take any point
E in the one, and the point F in the other: The straight line
which joins E and F is in the same plane with the parallels.

If not, let it be, if possible, above the plane, as EGF;
and in the plane ABCD in A
which the parallels are, draw
the straight line EHF from É
to F; and since EGF also is a

H
straight line, the two straight
lipes EHF, EGF include a
space between them, which is a

D
impossible. Therefore the straight line joining the points 10 Ax. 1.
E, F is not above the plane in which the parallels AB, CD
are, and is therefore in that plane. Wherefore, if two
straight lines, &c. Q.E.D.

PROP. VIII. THEOR.

If two straight lines be parallel, and one of them See N.
is at right angles to a plane; the other also shall
be at right angles to the same plane.

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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 joia *7.11. 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, AĚ, 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 equal 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
equal to the angle EDB, because each

C
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 B
base AE is common to the triangles

ABE, EDA; wherefore the angle
8. 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 3 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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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.

a 4. 11.

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 EF, 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 H
and GK, EF is perpendiculara to

FB
the plane HGK passing through
them: and EF is parallel to AB;
therefore AB is at right angles b E

G

F. 8. 11. to the plane HGK. For the same reason, CD is likewise at right angles to the plane HGK. There- K 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 parallel to one another. There- . 6. 11. fore AB is parallel to CD. Wherefore, two straight lines, &e. Q.E.D.

PROP. X. THEOR.

If two straight lines meeting one another be paral
lel 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

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• 33. 1.

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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

B
and parallel to BE. Therefore AD and
CF are each of them equal and parallel
to BE. But straight lines that are pa-

А.
rallel to the same straight line, and not
9. 11. in the same plane with it, are parallel

to one another. Therefore AD is paralo | Ax. 1. lel to CF; and it is equalc to it, and

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

D
are equal to DE, EF, and the base AC

F d 8. 1. to the base DF; the angle ABC is equald to the angle

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

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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 draw a AD perpendicular to BC. If thien AD be

also perpendicular to the plane BH, the thing required is already done; but if it be not,

А b 11. 3. from the point D drawb, in

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

H
pendicular to DE; and through
* 31. 1. F drawc GH parallel to BC:

And because BC is at right an

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

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 ane 8. 11. gles to a plane, the other shall be at right e angles to the

same plane; wherefore GH is at right angles to the plane f3 Def. 11. through ED, DA, and is perpendicularf to every straight

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

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plane through ED, AD, meets it: Therefore GH is per- Boox 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.

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 D

B at right angles to the plane.

From any point B above the plane draw a BC perpendicular to it; and

a 11. 11. from A drawb AD parallel to BC. Because, therefore, AD, CB are two parallel straight lines, and one of them BC is at right angles to the

А С given plane, the other AD is also at right angles to it c: 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.

31.1.

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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

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