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

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

Let AB, CD, meet the plane in the points B, D, and join 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 stright 2. 3. Def.s1. line which meets it, and is in that plane". 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, b. 29. 1.

CDB are together equal o 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 of them is a right 0. 4. .

angle; therefore the base AD is equal o
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
base AE is common to the triangles ABE, B

D IDA; wherefore the angle ABE is equal d. 8. 1.

a to the angle EDA. and ABE is a right
angle; and therefore EDA is a right angle,

E
and ED perpendicular to DA. but it is
also perpendicular to BD; therefore ED

is perpendicular e to the plane which passes thro' BD, DA, and 1.3.Def.14. fhall : make right angles with every straight line meeting it in that

plane. but DC is in the plane passing thro' 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 angles to the plane paffing thro' DE, DB, which is the fame plane to which AB is at right angles. Therefore if two straight lines, &c. Q. E. D.

e. 4. Ir.

197

Bock XI.

PROP. IX. THEO R.

TWO
"WO 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. Il.

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

In EF take any point G, from which draw, in the plane paffing thro' EF, AB, the straight line GH at right angles to EF; and in the plane passing thro' EF, CD, draw GK at right angles to the faine EF. and because EF is per- A H pendicular both to GH and GK, EF

B is perpendicular to the plane HGK passing thro' them. and EF is parallel

G to AB; therefore AB is at right an

F gles b to the plane HGK. for the fame reafon, CD is likewise at right angles to the plane HGK. therefore CK D AB, CD are each of them at right angles to the plane HGK. but if two straight lines be at right angles to the fame plane, they Mall be parallelo to one another. therefore AB is parallel to CD. c. 6. 11, Wherefore two straight lines, &c. Q. E. D.

b. 8. 11.

PROP. X. THEO R.

IF two straight lines meeting one another be parallel

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 another be parallel to the two straight lines DE, EF that meet one another, and are not in the faine plane with AB, BC. The angle ABC iş equal to the angle DEF.

Take BA, BC, ED, EF all equal to one another; and join AD, CF, BE, AC, DF. because BA is equal and parallel to ED, there;

Book XI. fore AD is both equal and parallel to

BE. For the same reafon, CF is equal 2. 33. I.

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

А

C to BE. Bat straight lines that are paral

iel to the fame straight line, and not in b. 9. 11. the fame plane with it, are parallel b to

one another. Therefore AD is parallel 6. 1. Ax. 1. to CF; and it is equal to it, and AC, DI join them towards the fame parts ; D

F and therefore, a AC is equal and parallel

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

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

PRO P. XI. PROB.

T.

o draw a straight line perpendicular to a plane, from a given point above it.

Let A be the given point above the plane BII; it is required to draw froin the point A a straight line perpendicular to the plane BH.

In the plane draw any straight line BC, and from the point A draw a AD perpurdicular to LC. If then AD bu als perpendi

cular to the plane DH, the thing required is alrcady done, but if it b. 11.1.

he not, from the point D draw b
in the plane BH, the straight line
DEat right angles to BC; and from
the point A dawa perpendicular Gc

H
to DE; and thro' F draw o CH pa-
rallel to BC. and because BC is at

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

fing thro' ED, DA. And GII is 3 D
parallel to BC; but if two straight lines be parallel, one of which is
at right angles to a plane, the other fhall be at right © angles to

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

that pline. But AF, which is in the plane thro’ED, DA meets it,

C. 31. I.

e. 8. 11.

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.

PROP. XII. PROB.

To crect a ftraight line at right angles to a given

plane, from a point given in the plane.

2. II. II.

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
gles to the plane.

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

b. 3. 1. A draw 6 AD parallel to BC. because therefore AD, CB are two parall straight A lines, and one of them BC is at right angles to the given plane, the other AD ish also at right angles to it. therefore a straight line has been erect. c. 8.11, ed at right angles to a given plane from a point given in it. Which was to be done.

PRO P. XIII. THEOR.

FRO

*ROM 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 pafs thro' BA, AC; the common section of this with the given plane is a straight a line

3. !!!

Book XI. passing through A. let DAE be their common fection. therefore

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

B C
that plane. but DAE which is in
that pline meets CA; therefore CAE
is a righe 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 E which is imposible. Also, from a point above a plane there can

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

fore from the same point, &c. 0. E. D.

PROP. XIV. THEOR.

PLA

LANES to which the same straight line is perpendicular, are parallel to one another.

к

Let the straight line AB be perpendicular to each of the planes CD, EF; these planes are parallel to one another.

If not, they hall meut one onother 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 be

cause AB is perpendicular to the plane 4.3. Def. 11. EF, it is perpendicular a to the straight C

H line BK which is in that plane. there

F fore ADK is a right angle. for the fame A

B reason, BAK is a right angle; wherefore the two angles AEK, BAK of the triangle ABK are equal to two right

E b. 17. I. angles, which is imposible b. therefore

D the planes CD, EF though produced do c.8. Def.11. not meet one another ; that is, they are parallel Therefore

planes, &c. Q: E. D.