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Book XI.

Let AB, CD be two parallel ftraight lines, and let one of them AB be at right angles to a plane; the other CD is at right angles to the fame 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 ftraight line which a. 3. Def.11. 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 Farallel straight lines AB, CD, the angles ABD, CDB are together b. 29. I. equal to two right angles. and ABD is a right angle; therefore allo 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, becaufe each A

of them is a right angle; therefore the é. 4. 1. bafe AD is equal to the base BE. again,

c

because AB is equal to DE, and BE to

AD; the two AB, BE are equal to the

two ED, DA; and the bafe AE is com- B
mon to the triangles ABE, EDA; where-

d

d. 8. 1. fore the angle ABE is equal to the angle

EDA. and ABE is a right angle; and

therefore EDA is a right angle, and ED

e

E

D

perpendicular to DA. but it is alfo perpendicular to BD; therefore 4.11. ED is perpendicular to the plane which paffes through BD, DA, £3.Def.11.and shall f make right angles with every straight line meeting it in

that plane. but DC is in the plane paffing through BD, DA, becaufe 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 sight angles to DE. but CD is alfo at right angles to DB; CD then is at right angles to the two ftraight lines DE, DB in the point of their interfection 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.

PROP

Book XI.

PROP. IX. THEOR.

TWO ftraight lines which are each of them parallel to the fame ftraight line, and not in the fame plane

with it, are parallel to one another.

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 passing thro' EF, AB, the straight line GH at right angles to EF; and in the plane paffing through EF, CD, draw GK at right angles to the fame EF. and because EF is perpen

dicular both to GH and GK, EF is

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A H

E

B

2. 4. II.

G

F

b. 8.11.

D

perpendicular to the plane HGK paffing through them. and EF is parallel to AB; therefore AB is at right angles b to the plane HGK. for the fame reason, CD is likewife at right angles to the plane HGK. therefore AB, CD are each of them at right angles to the plane HGK. but if two ftraight lines be at right angles to the fame plane, they fall be parallel to one another. therefore AB is parallel to CD. Where- c. 6. 11. fore two straight lines, &c. Q. E. D.

IF

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PROP. X. THEOR.

two straight lines meeting one another be parallel to two others that meet one another, and are not in the fame plane with the first two; the first two and the other two fhall 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 fame plane with AB, BC. the angle ABC is equal to the angle DEF.

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

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Book XI. CF, BE, AC, DF. because BA is equal and parallel to ED, there

fore AD is both equal and parallel to BE.

a. 33. 1. for the fame reafon, CF is equal and pa

rallel to BE. therefore AD and CF are

each of them equal and parallel to BE. but A
straight lines that are parallel to the fame
ftraight line, and not in the fame plane

D

B

C

E

F

b. 9. 11. with it, are parallel b to one another. therec. 1. Ax... fore AD is parallel to CF; and it is equal to it, and AC, DF join them towards the fame parts; and therefore AC is equal and parallel to DF. and becaufe AB, BC are equal to DE, EF, and the base AC to the base DF; the angle ABC is equal to the angle DEF. Therefore if two straight lines, &c. Q. E. D.

d. 8. 1.

To dr

PROP. XI. PROB.

O 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 point A 2. 12. 1. draw AD perpendicular to BC. If then AD be alfo perpendicular to the plane BH, the thing required is already done; but if it be

a

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rallel to BC; but if two ftraight lines be parallel, one of which is

e. 8. 11. at right angles to a plane, the other fhall be at right angles to the fame plane; wherefore GH is at right angles to the plane thro'

f. 3. Def. 11. ED, DA, and is perpendicular f to every ftraight 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 confequently 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 interfection, it shall also be at right angles to the plane paffing through them. but the plane paffing 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 ftraight 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 angles to the plane.

From any point B above the plane draw BC perpendicular to it; and from A draw AD parallel to BC. because therefore AD, CB are two parallel straight lines, and one of them BC is at right angles to the given plane, the other AD is alfo at

D

B

a. II. II.

b. 31. 1.

Α

'C

right angles to it. therefore a ftraight line has been erected at right c. 8. 11. angles to a given plane from a point given in it. Which was to be done.

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FROM the fame point in a given plane there cannot be two ftraight lines at right angles to the plane, upon the fame fide of it. and there can be but one perpendicular to a plane from a point above the plane.

For, if it be poffible, let the two ftraight lines AB, AC be at right angles to a given plane from the fame point A in the plane, and upon the fame fide of it; and let a plane pass through BA, AC; the common fection of this with the given plane is a straight line a. 3. 11.

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Book XI. paffing through A. let DAE be their common fection. therefore the ftraight 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 that
plane meets CA; therefore CAE is
a right angle. for the fame reafon
BAE is a right angle. wherefore the
angle CAE is equal to the angle
BAE; and they are in one plane,

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which is impoffible. Alfo, 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 b to one another, which is abfurd. Therefore from the fame point, &c. Q. E. D.

PROP. XIV. THEOR.

PLANES to which the fame straight line is perpendicular, are parallel to one another.

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

If not, they fhall meet one another when produced; let them meet; their common fection shall be a

G

ftraight line GH, in which take any

point K, and join AK, BK. then be-
caufe AB is perpendicular to the plane

a. 3.Def.11. EF, it is perpendicular to the straight C
line BK which is in that plane. there-
fore ABK is a right angle. for the fame
reason, BAK is a right angle; where-
fore the two angles ABK, BAK of the
triangle ABK are equal to two right

b. 17. 1. angles, which is impoffible b. therefore

the planes CD, EF tho' produced do

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c. 8.Def.11. not meet one another; that is, they are parallel. Therefore

planes, &c. Q. E. D.

PROP.

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