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 reason 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;

b. 4. I.

F

d. 8. r.

C

A

and the base DF was proved equal to the
base FC; therefore the angle FAD is e-
qual d to the angle FBC. again, it was
proved that AG is equal to BH, and alfo
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 base GF is equal to the
base FH. again, because it was proved
that GE is equal to EH, and EF is com-D
mon; GE, EF are equal to HE, EF; and

G

the base GF is equal to the base FH; therefore the angle GEF

is equal d to the angle HEF, and consequently each of these angles is a right angle. Therefore FE makes right angles with GH, that 6.10. Def.1.

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.

I

PROP. V. THEOR.

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

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

Let the straight 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 possible, BD and BE be in one plane, and BC be above it; and let a plane pass through AB, BC, the common section of which with the plane, in which BD and BE are, shall be a

N

a. 3. 11.

Book XI. straight a line; let this be BF. therefore the three straight lines AB, BC, BF are all in one plane, viz. that which passes thro' AB, BC. and because ABstands at rightangles toeach of the straight lines BD, BE, it is also at right angles to the plane passing thro' them; and there-A

b. 4. 11.

C

F

D

c.3.Def.11. fore makes right angles with every straight 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 Hypothesis, is also a right angle; therefore the angle ABF is equal to the angle ABC, and they are both in the same plane, which is impossible. 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 same plane. Therefore if three straight lines, &c. Q. E. D.

I

B

PROP. VI. THEOR.

E

F two straight lines be at right angles to the fame plane, they shall 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 straight 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 shall make 2.3.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, right angle. and because AB is equal to DE, and BD common, the two

CDE is

a

C

D

fides AB, BD, are equal to the two ED, DB; and they contain right angles; therefore the base AD is equal to the base BE. again, because AB is equal to DE, and BE to AD; AB, BE are equal to ED,

:

Book XI.

c. 8. 1.

DA, and, in the triangles ABE, EDA, the base AE is common; 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 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 fame plane d. but AB d. 5. 11. is in the plane in which are BD, DA, because any three ftraight lines which meet one another are in one plane e. therefore AB, 2. 2. 11. 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.

I

E. D.

PROP. VII. THEOR.

14

F two straight lines be parallel, the straight line drawn See N. 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 fame plane with the parallels.

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

lels are, draw the straight line EHF AE

1

B

plane in which the parallels AB, CD are, and is therefore in that

plane. Wherefore if two straight lines, &c.

PROP. VIII. THEOR.

E. D.

F two straight lines be parallel, and one of them is Sce N.

IF

at right angles to a plane; the other also shall be at

right angles to the fame plane.

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 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 8.3. 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,

b. 29. 1.

is perpendicular to the plane which passes thro' BD, DA, and f. 3. Def.11. shall f 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 interfection D; and therefore is at right angles to the plane passing thro' DE, DB, which is the same plane to which AB is at right angles. Therefore if two straight lines, &c. Q. E. D.

TWO

PROP. IX. THEO R.

WO straight lines which are each of them parallel
to the same straight 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 thro' EF, CD, draw GK at right angles to the

:

Book XI.

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 shall be parallel to one another. therefore AB is parallel to CD. c. 6. 11.

Wherefore two straight lines, &c. Q. E. D.

IF

PROP. X. THEOR.

F 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 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 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