d

T

b. 4. I.

equal to EH, and AG to BH. and because AE is equal to EB, and Book XI. FE common and at right angles to them, the base AF is equal to the base FB; for the fame reafon CF is equal to FD. and because AD is equal to BC, and AF to FB, the two fides FA, AD are equal to the two FB, BC, each to each; and the base DF was proved equal to the bafe FC; therefore the angle FAD is equal to the angle FBC. again, it was proved that AG is equal to BH, and also AF to FB; FA then and AG, are equal to FB and BH, and the angle FAG has been proved equal to the angle FBH; therefore the bafe GF is equal to the bafe FH. again, because it was proved that GE is equal to EH, and EF is common; GE, EF are equal to HE, EF; and

d. S. 1.

C

E

H

B

the bafe GF is equal to the bafe FH; therefore the angle GEF is equal to the angle HEF, and confequently each of these angles is a right angle. Therefore FE makes right angles with GH, that e.10.Def.. is, with any straight line drawn thro' E in the plane passing thro' AB, CD. In like manner it may be proved that FE makes right angles with every straight line which meets it in that plane. But a straight line is at right angles to a plane when it makes right angles with every straight line which meets it in that plane f. therefore EF f. 3. Def.11. is at right angles to the plane in which are AB, CD. Wherefore if a straight line, &c. Q. E. D.

F three ftraight lines meet all in one point, and a see N. I ftraight line ftands at right angles to each of them

in that point; these three ftraight lines are in one and the fame plane.

Let the ftraight line AB stand at right angles to each of the straight lines BC, BD, BE, in B the point where they meet; BC, BD, BE are in one and the fame plane.

If not, let, if it be poffible, BD and BE be in one plane, and BC be above it; and let a plane pass through AB, BC, the common fection of which with the plane, in which BD and BE are, fhall be a

N

Book XI. ftraight a line; let this be BF. therefore the three straight lines AB, BC, BF are all in one plane, viz. that which paffes thro' AB, BC. and because AB ftands at right angles to each of the ftraight lines BD, b. 4. II. BE, it is alfo at right angles to the

a. 3. II.

b

plane passing thro' them; and there-A c.3.Def.11. fore makes right angles with every

C

F

D

B

E

ftraight line meeting it in that plane;
but BF which is in that plane meets
it, therefore the angle ABF is a right
angle; but the angle ABC, by the
Hypothefis, is also a right angle;
therefore the angle ABF is equal to
the angle ABC, and they are both
in the fame plane, which is impoflibfe. therefore the straight line
BC is not above the plane in which are BD and BE. wherefore
the three straight lines BC, BD, BE are in one and the fame plane.
Therefore if three straight lines, &c. Q. E. D.

F

IT

two ftraight lines be at right angles to the fame plane, they fhall be parallel to one another.

Let the straight lines AB, CD be at right angles to the fame plane; AB is parallel to CD.

Let them meet the plane in the points B, D, and draw the ftraight line BD, to which draw DE at right angles, in the fame plane; and make DE equal to AB, and

join BE, AE, AD. Then because AB is A perpendicular to the plane, it fhall make a.3.Def.11. right a angles with every straight line which meets it, and is in that plane. but BD, BE, which are in that plane, do each of them meet AB. therefore each of the B angles ABD, ABE is a right angle. for the fame reason, each of the angles CDB,

b. 4. I.

D

E

CDE is a right angle. and because AB is
equal to DE, and BD common, the two
fides AB, BD, are equal to the two ED, DB; and they contain right
angles; therefore the base AD is equal to the bafe BE. again, be-
cause AB is equal to DE, and BE to AD; AB, BE are equal to ED,

per

c. 8. I.

DA, and, in the triangles ABE, EDA, the bafe AE is common; Book XI. therefore the angle ABE is equal to the angle EDA. but ABE is a right angle; therefore EDA is also a right angle, and ED pendicular to DA. but it is alfo 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 ftraight lines are all in the fame plane 4. but AB d. 5. 11. is in the plane in which are BD, DA, because any three straight lines which meet one another are in one plane e. therefore AB, BD, DC are in one plane. and each of the angles ABD, BDC is a right angle; therefore AB is parallel f to CD. Wherefore if f. 28. 1. two straight lines, &c. Q. E. D.

PROP. VII. THEOR.

e. 2. II.

If two line other

F two ftraight lines be parallel, the ftraight line drawn See N.

is in the fame plane with the parallels.

Let AB, CD be parallel ftraight 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 fame plane with the parallels.

If not, let it be, if poffible, above the plane, as EGF; and in the

a. to. Ax.1.

F D

twixt them, which is impoffible a. Therefore the straight line joining C

the points 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.

IF

PROP. VIII. THEOR.

two straight lines be parallel, and one of them is See N. at right angles to a plane; the other alfo fhall be at right angles to the fame plane.

Book XI.

b. 29. I.

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 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 straight B.3.Def.11. 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, 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 of them is a right angle; therefore the base AD is equal to the bafe 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 bafe AE is common to the triangles ABE, EDA; wherefore 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 alfo perpendicular to BD; therefore ED

C. 4. I.

d. 8. 1.

e. 4. 11.

e

A

B

is perpendicular to the plane which paffes thro' BD, DA, and f. 3. Def.11. shall ƒ make right angles with every straight line meeting it in that plane. but DC is in the plane paffing 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 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. IX. THEOR.

WO 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 fhall be parallel to CD.

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

pendicular both to GH and GK, EF

a

Book XI.

B

is perpendicular 2 to the plane HGK paffing thro' them. and EF is parallel

2. 4. II.

G

to AB; therefore AB is at right an

E

-F

gles to the plane HGK. for the

b. 8. 11.

D

fame reason, CD is likewise at right

angles to the plane HGK. therefore C K

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

fhall be parallel to one another. therefore AB is parallel to CD. c. 6. 12. Wherefore two ftraight lines, &c. Q: E. D.

PROP. X. THEOR

[F two ftraight lines meeting one another be parallel

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