Let the straight line AB be at right angles to a plane CK; Book II. every plane which paffes through AB is at right angles to the plane CK.

Let any plane DE pafs through AB, and let CE be the common section of the planes DE, CK; take any point F in CE, from which draw FG in the plane DE at right angles to CE: And because AB is perpendicular to the plane CK, therefore it is alfo perpendicular

to

every

ftraight line meeting it in that plane a; and confequently it is perpendicular to CE: Wherefore ABF

is a right angle; but GFB is likewife a right angle; therefore AB is parallel

D

G

A

H

But one

c 7. 2. Sup.

to FG. And AB is at right angles to the plane CK; therefore FG is also at right angles to the fame plane c. plane is at right angles to another plane when the straight lines drawn in one of the planes, at right angles to their common section, are also at right angles to the other planed; and a 2. def. 2. any straight line FG in the plane DE, which is at right angles to CE, the common fection of the planes, has been proved to be perpendicular to the other plane CK; therefore the plane DE is at right angles to the plane CK. In like manner, it may be proved that all the planes which pass through AB are at right angles to the plane CK. Therefore, if a ftraight line, &c. Q.E. D.

Let the two planes AB, BC be each of them perpendicular to a third plane, and let BD be the common fection of the first two; BD is perpendicular to the plane ADC.

R 4

From

Supplement From D in the plane ADC, draw DE perpendicular to
'AD, and DF to DC. Becaufe DE is perpendicular to AD,
the common fection of the planes
AB and ADC; and because the plane
AB is at right angles to ADC, DE

a 2. def. 2. is at right angles to the plane AB,
Sup. and therefore alfo to the ftraight line
b 1. def. 2. BD in that plane b. For the fame rea-
Sup.
fon, DF is at right angles to DB.
Since BD is therefore at right angles
to both the lines DE and DF, it is at
right angles to the plane in which DE
and DF are, that is, to the plane

c4. 2. Sup. ADC c. Wherefore, &c. Q. E. D.

B

AF

E

C

TW

WO ftraight lines not in the fame plane being given in pofition, to draw a ftraight line perpendicular to them both.

Let AB and CD be the given lines, which are not in the fame plane; it is required to draw a straight line which shall be perpendicular both to AB and CD.

In AB take any point E, and through E draw EF parallel to CD, and let EG be drawn perpendicular to the plane which paffes through

EB, EF.

Through

G

B

b cor. 13.2.

in the fame plane, are both at right angles to the ftraight line Book II. AB, they are parallel to one another. And because the lines HG, HD are parallel to the lines EB, EF, each to each, the plane GHD is parallel to the plane a BEF; and therefore a 9. 2. Sup. EG, which is perpendicular to the plane BEF, is perpendicular alfo to the plane b GHD. Therefore HK, which is rallel to GE, is also perpendicular to the plane GHD, and Sup. it is therefore perpendicular to HD 4, which is in that plane, Sup. and it is alfo perpendicular to AB; therefore HK is drawn d 7. 2. Sup. perpendicular to the two given lines, AB and CD. Which was to be done.

pa

c def. i. 2.

F a folid angle be contained by three plane angles, any two of them are greater than the third.

i

Let the folid angle at A be contained by the three plane angles BAC, CAD, DAB. Any two of them are greater than the third.

D

If the angles BAC, CAD, DAB be all equal, it is evident that any two of them are greater than the third. But if they are not, let ABC be that angle which is not less than either of the other two, and is greater than one of them DAB; and at the point. A in the ftraight line AB, make, in the plane which paffes through BA, AC, the angle BAE equal to the angle DAB; and make AE equal to AD, and through E draw BEC cutting AB, AC in the points B, C, and join DB, DC. And because DA is equal to AE,

a

C

B

E

a 23.

and AB is common to the two triangles ABD, ABE, and also the angle DAB equal to the angle EAB; therefore the bafe DB is equal to the bafe BE. And be- b 4. I, cause BD, DC are greater than CB, and one of them BDC 20. I. has been proved equal to BE, a part of CB, therefore the other DC, is greater than the remaining part EC. And because DA is equal to AE, and AC common, but the base

DC

250

d 25. 1.

Supplement DC greater than the base EC; therefore the angle DAC is greater d than the angle EAC; and, by the construction, the angle DAB is equal to the angle BAE; wherefore the angles DÅB, DAC are together greater than BAE, EAC, that is, than the angle BAC. But BAC is not lefs than either of the angles DAB, DAC; therefore BAC, with either of them, is greater than the other. Wherefore, if a folid angle, &c. Q. E: D.

THE

PROP. XXI. THEOR.

HE plane angles, which contain any folid angle, are together less than four right angles.

Let A be a folid angle contained by, any number of plane angles BAC, CAD, DAE, EAF, FAB; thefe together are lefs than four right angles.

Let the planes which contain the folid angle at A be cut by another plane, and let the section of them by that plane be the rectilineal figure BCDEF. And because the folid angle at B is contained by three plane angles CBA, ABF, FBC, of which a 20. 2.Sup. any two are greater a than the

b 32. I.

CI. cor. 32. I.

B

A.

E

third, the angles CBA, ABF are
greater than the angle FBC: For
the fame reafon, the two plane
angles at each of the points C, D,
E, F, viz. the angles which are at
the bafes of the triangles having
the common vertex A, are great-
er than the third angle at the fame
point, which is one of the angles
of the figure BCDEF: there-
fore all the angles at the bases of the triangles are together
greater than all the angles of the figure and because all
the angles of the triangles are together equal to twice as ma-
ny right angles as there are triangles b; that is, as there are
fides in the figure BCDEF; and becaufe all the angles of
the figure, together with four right angles, are likewife
equal to twice as many right angles as there are fides in the
figure; therefore all the angles of the triangles are equal to

D

all

all the angles of the rectilineal figure, together with four right Book II. angles. But all the angles at the bafes of the triangles are greater than all the angles of the rectilineal, as has been proved. Wherefore, the remaining angles of the triangles, viz. those at the vertex, which contain the folid angle at A, are lefs than four right angles. Therefore every folid angle, &c. Q. E. D.

Otherwise.

Let the fum of all the angles at the bafes of the triangles =S; the fum of all the angles of the rectilineal figure BCDEF=Σ; the fum of the plane angles at A=X, and let R a right angle.

Then, because S+X twice b as many right angles as there b 32. 1. are triangles, or as there are fides of the rectilineal figure BCDEF, and as Σ + 4R is alfo equal to twice as many right angles as there are fides of the fame figure; therefore S+X=E+4R• But because of the three plane angles which contain a folid angle, any two are greater than the third, S> Σ; and therefore X <4R; that is, the fum of the plane angles which contain the solid angle at A is lefs than four right angles. Q. E. D.

SCHOLIUM.

It is evident, that when any of the angles of the figure BCDEF is exterior, like the angle at D, in the annexed figure, the reasoning in the above propofition does not hold, because the folid angles. at the base are not all contained by plane angles, of which two belong to the inclined planes of the triangles, and the third is an interior angle of the rectilineal figure,

B

D

or bafe. Therefore, it cannot be concluded that S is, neceffarily, greater than E. This propofition, therefore, is fubject to a limitation, which is farther explained in the notes on this book.

ELE