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

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 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 perpendi

cular to the plane, it is perpendicular to every straight line which a. 3. Def.nl.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, CDB are together b. 29. 1. equalb 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, because each A

of them is a right angle; therefore the 6. 4. 1. base AD is equal to the base BE. again,

because AB is cqual to DE, and BE to
AD; the two AB, BE are equal to the
WED, DA; and the base AE is com- D

. inon to the triangles ABE, EDA; whered. 8. 1. fore the angle ABE is equal d to the angle

EDA. and ABE is a right angle ; and
therefore EDA is a right angle, and ED

perpendicular to DA. but it is also perpendicular to BD; therefore € 4.11. ED is perpendiculare to the plane which passes through BD, DA, 1.3.Def.11. and fhall 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 yight angles to DE. bui CD is also at right angles to DB; CD thenis at right angles to the two straight lines DE, DB in the point of their interfection D; and therefore is at right angles to the plane paling thro' DE, DB, which is the same plane to which AB is at right angles. Therefore if two straight lines, &c. Q. E. D.

PROP

Book XI.

PROP. IX.

THEOR.

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.

2.4. II.

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 paling thro' 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 fame EF. and because EF is perpen

A H dicular both to GH and GK, EF is

-B perpendicular to the plane HGK passing through them. and EF is pa

G rallel to AB; therefore AB is at right angles b to the plane HGK. for the

b. 8.11. fame reason, CD is likewise at right

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D angles to the plane HGK. therefore AB, CD are each of them at right angles to the plane HGK, but if two straight lines be at right angles to the same plane, they will be parallel o to one another. therefore AB is parallel to CD. Where. c. 6.11. fore two straight lines, &c. Q. E. D.

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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 iwo 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 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 join AD,

- CF,

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

B 2. 33. 1. for the same reason, CF is equal and pa

rallel to BE. therefore AD and CF are
each of them equal and parallel to BE, but A

С
straight lines that are parallel to the same

straight line, and not in the same plane b. 9.18. with it, are parallel b to one another, therec. 1. Ax..1. fore AD is parallel to CF; and it is equal o to it, and AC, DF join them towards the

E fame parts ; and therefore a AC is equal

D and parallel to DF. and because AB, BC

are equal to DE, EF, and the base AC to the base DF; the angle d. 8. 1.

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

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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 point A 2. 12. 1. draw a AD perpendicular to BC. If then AD be also perpendicular

to the plane BH, the thing required is already done; but if it be b. 11. 1. not, from the point D draw bin the

A
plane BH, the straight line DE at
right angles to BC; and from the

E
point A draw AF perpendicular to G

F

H C. 31. 1. DE; and thro' F draw o GH pa

rallel to BC. and because BC is at

right angles to ED and DA, BC is d. 4. 11. at right anglesd to the plane palling through ED, DA. and GH is pa

B
D

с rallel to BC; but if two straight lines be parallel, one of which is e. 8. 11. at right angles to a plane, the other shall be at right ° angles to

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

ΤΟ

erect a straight line at right angles to a given plane, from a point given in the plane.

a. 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

D B angles to the plane.

From any point B above the plane draw BC perpendicular to it; and from A draw 6 AD parallel to BC. because there

b. 31. I. fore AD, CB are two parallel straight lines, and one of them BC is at right angles

'C to the given plane, the other AD is also at right angles to it. therefore a straight 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 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 pass through BA, AC; the common section of this with the given plane is a straight a line a. 3. 11.

pafling

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

B С
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. wherefore the
angle CAE is equal to the angle
BAE; and they are in one plane,

D А
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 b to one another, which is absurd. Therefore

from the same point; &c. Q. E. D.

PROP. XIV.

THEOR.

PLANES to which the same straight 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 thall 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 TI

line BK which is in that plane. there-
fore ABK is a right angle. for the same

A А

B
reason, BAK is a right angle ; where-
fore the two angles ABK, BAK of the
triangle ABK are equal to two right

E b. 17. 1. angles, which is impossible b. therefore

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

planes, &c. Q. E. D.

PROP.

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