angle specified in the enunciation. In like manner may the other dihedral angles be shown to be equal.

THEOREM XXV. (126.)

If two solid angles be contained, each by three plane angles which are equal to one another, each to each, and follow each other in the same order, these solid angles are equal.

LET there be two solid angles at A and B, contained by the three plane angles CAD, DAE, EAC, taken in order, and FBH, HBG, GBF, also taken in the same order, such that CAD = FBH, DAE = HBG, and EAC = GBF. Then will the solid angle at A be equal to the solid angle at B.

E

F

Let the solid angle at A be applied to the solid angle at B, so that the plane angle CAD coincide with the plane angle FBH. Then since CA coincides with BF, and the dihedral angle made by CAD, CAE, is equal to the dihedral angle made by FBH and FBG (th. 125), the plane CAE will coincide with the plane FBG. Also, since the angles CAE, FBG, are equal, the line AE coincides with BG. Wherefore the plane angle EAD coincides with the plane angle GBH; and the solid angles at A and B thus coinciding, are equal to one another.

THEOREM XXVi. (127.)

If every two of three plane angles be greater than the third, and if the straight lines which contain them be all equal, a triangle may be made of the straight lines which join the extremities of those containing lines.

GHK, and BL equal either of the equal lines AB, &c. Join AL, LC.

C

Then the triangles GHK, ABL, are equal, and therefore AL = GK. But the angles DEF and GHK, being greater than ABC, and GHK = ABL, therefore DEF is greater than LC, and the base DF is greater than LC. But AL and LC are together greater than AC; much more then are AL and DF (that is, GK and DF) together greater than AC. In the same way it may be shown, that ACDF are greater than GK, and AC + GK greater than DF. Hence of the three lines AC, DF, GK, any two are greater than the third; and therefore a triangle can be constituted of them.

Cor. If from the three edges of a triangular pyramid a perpendicular be drawn to the plane of the base, it will meet the base in the centre of the circle which circumscribes the base. See the next figure.

THEOREM XXVII.

If about a triangle ABC a circle be described, and from its centre E a perpendicular ED be drawn perpendicular to the plane of the base, any point D in this perpendicular will be the vertex of a triangular pyramid, whose three edges are equal.

LET the three edges DA, DB, DC, of the tetrahedron DABC be all equal; and from D draw DE perpendicular to the plane ABC of the base. Then E will be the centre of a circle described about the triangular base ABC.

For join EA, EB, EC. Then since DE is perpendicular to ABC, the three angles DEA, DEB, DEC, are right angles.

D

=

B

EC2.

Whence AD2 — DE2 = EA2, BD2 - DE2 = EB2, and CD2 - DE2 But by hypothesis AD2 = BD2 = CD2, and hence also AE = BE CE; and E is the centre of the circle passing through the points A, B, C.

THEOREM XXVIII.

If two solid angles be contained each by three plane angles which are equal two and two: then in one solid angle any edge will have the same inclination to the plane of the two others, that the homologous edge of the other has to the homologous plane of the other two.

LET the solid angles A and B be contained by plane angles CAE, EAD, DAC, and FBH, HBG, GBF, which are two and two equal respectively; then the edge AC will be inclined to the plane EAD in the same angle that the homologous edge BF is to the homologous plane HBG.

F

D

P

G

R

K B

L

E

H

Take AC equal to BF, and draw the perpendiculars CK, FL, to the planes DAE, GBH, meeting them in K, L; and from K, L, draw KP, LR, perpendicular to the homologous lines AD, BG; and join CP, FR.

Then (th. 102) CP is perpendicular to AD, and FR to BG. Hence CPK, FRL, are the inclinations of the planes CAD, DAE, and FBG, GBH, respectively; and these inclinations (th. 125) are equal to one another, and therefore the angles CPK, FRL, are equal.

Again, since in the right-angled triangles CAP, FBR, the homologous sides AC, BF, are equal, and the angles CAD, FAG, also equal, the sides CP, FR, are also equal. Whence in the right-angled triangles CKP, FLR, the homologous sides CP, FR, are equal, and the angles CPK, FRL, also equal, the side CK is equal to the side FL.

Hence, we have in the right-angled triangles CKA, FLB, the two sides AC, CK, equal to the two BF, FL, each to each, the angle CAK is equal to FBL: and these are the inclinations of AC to DAE, and of FB to GBH. Hence these inclinations are equal.

SECTION III.-THE VOLUMES OF SOLIDS *.

THEOREM XXIX. (114.)

If a prism be cut by a plane parallel to the ends, it will be divided in the same ratio as any one of its parallel edges is divided by the same plane.

LET the plane EF cut the prism AD into two parts AF, ED, and any one of its parallel edges AC into the parts AE, EC: then

prism AF : prism ED :: AE : EC.

W

U

F D X Y Z

L

IM

T

H K

For produce the planes AK, AN, BN, BK; and in the produced edge AC take any number of lines AO, OP, each equal to AE, and any number CQ, QR, RS, each equal to EC, and through the several points O, P, Q, R, S, let planes OU, PW, QX, RY, SZ, be drawn parallel to AB, CD, or EF.

Then since TO is parallel to AG, and OV parallel to AL, the three plane angles TOV, VOA, AOT, are equal to the three GAL, LAE, EAG, each to each. Hence if the point O were applied to A, the line OA to the line AE, and the plane AT to the plane EG, the line OT would coincide with the line AG, and the line OV with the line AL.

For a similar reason, the lines TU, UV, UB, would coincide with GB, BL, BF; the lines BL, BG, with FM, FH, and TG, GA, with GH, HE. Hence, all the edges of the prism OB would coincide with all the homologous edges of the prism AF; and therefore all the faces of the one coincide with all the faces of the other, and the two prisms are equal.

In the same manner, it may be proved that OW is equal to AF; and similarly, that each of the prisms CX, QY, RZ, is equal to ED. Wherefore, whatever multiple the line PE is of the line AE, the same multiple is the prism EW of the prism EB; and whatever multiple the line ES is of the line CE, the same multiple is the prism EZ of the prism ED.

Again, if the line PE be greater than ES, the prism EW is greater than the prism EZ; if equal, equal; if less, less: and these are any equimultiples of the first and third AE, AF, and any equimultiples of the second and fourth EC, ED, each of each. Hence

prism AF: prism ED :: AE: EC.

* The following propositions are mainly proved by means of the method of Cavallerius, which he calls the arithmetic of infinites. It consists in assuming that all plane figures are made up of an infinite number of lines parallel to each other, and connected by a certain law according to the particular figure under consideration; and similarly, solids are assumed to be composed of an infinite number of indefinitely thin laminæ, or mere plane figures, all parallel to each other, and connected by the properties of parallel sections of the solid. This method is not rigorously conclusive: but the great length of the proofs by the method of exhaustions, renders them unsuitable to the space allowed to the subject in this course; and, as they all admit of a ready and brief investigation by means of the integral calculus, it does not appear to be essential to give a more rigid system of investigation in this place.

Scholium.

This demonstration being general for all prisms, the particular case of the rectangular parallopiped is included in it. The property as often used in reference to this particular figure is :-rectangular parallelopipeds of the same altitude are to one another as their bases.

THEOREM XXX. (113.)

Prisms and cylinders of the same altitude, or between the same parallel planes, are equal to one another.

LET ABT, DEV, FGXW, be prisms and a cylinder on equal bases ABC, DE, FG, respectively, and between the same parallel planes MN, PQ (or of the same altitude HI); they shall be equal to each other.

Parallel to the plane MN draw any plane M'N' cutting the prisms and cylinder in the sections A'B'C', D'E', F'G'. Then these are respectively equal to the sections ABC, DE, FG; and as these latter are equal by hypothesis, the former are also equal. But the prisms and cylinder being respectively made up of these equal laminæ, and equally numerous, they must also be equal.

P

R

M

A

M

F

Σ

B

D

Cor. Every prism and cylinder is equal to a rectangular parallelopiped, of equal base and altitude with it.

THEOREM XXXI. (115.)

Prisms of equal bases are to one another as their altitudes.

LET AL, ER, be two prisms on equal bases AC, EG; they shall be to one another as their altitudes TU, VW.

P

R

K

T

I

M

Q'

R

For, if the bases be not similar as well as equal, make the base EG similar to AC, and the prism ER upon it equiangular to the prism AL. Make AP' equal to EP, and draw the plane P'R' parallel to AC. Then the prism AR' is equal to ER; and since the prism AL is cut by the plane P'R' parallel to AC into parts proportional to AP', P'I, the whole prism AL: prism AR' :: AI : AP' :: UT: UX :: UT : VW :: altitude of prism AL: altitude of prism ER. That is, the prisms upon the equal bases AC, EG, are to one another as their altitudes.

THEOREM XXXII. (115. Cor. 2.)

Prisms, neither whose bases nor altitudes are equal, are to one another in the ratio compounded of the ratio of their bases and the ratio of their altitudes.

LET AL, EZ, be two prisms, neither of whose bases AC, EO, nor altitudes TU, VW, are equal; they will be to one another in the ratio compounded of the ratio of AC to EO, and the ratio of TU to VW.

P

I

耳

L

T

M

Q'

R

Z

R

Y

P

S

S

H

C

Le

For take AP' equal to EP, and draw the plane P'R' parallel to AC; and take EF equal to AB, and draw the plane FR parallel to EQ. Then the prism AR' will be equal to the prism ER'. Hence

prism ER base EG prism AL UX

= (th. 114, 115;) prism EZ base EO' prism AR TU'

whence, compounding these ratios, recollecting the equality of the prisms ER, AR', and of the altitudes UX, VW, we have

All pyramids and cones upon equal bases, and between parallel planes, or having equal altitudes, are equal.

LET the pyramids VABC, WDEF, and cone XGH, have equal bases ABC, DEF, GH, and have equal altitudes VR, WS, XT, or lie between the same parallel planes PQ, MN; they will be equal to each other.

For, draw any plane M'N' parallel to the plane in which their bases are situated, cutting them in the sections A'B'C', D'E'F', and G'H', respectively, and the perpendiculars VR, WS, XT, from their vertices to their bases in R', S', T'.

Then (th. 109) the perpendiculars are divided proportionally in R', S', T'; and (th. 117) the sections have to the bases the duplicate ratio of the perpendiculars VR', VR; and, therefore, since the bases are equal, the sections are equal.

Now this is true for all the sections that can be made parallel to the base, and hence for the sums of all such sections: but these taken together make up the entire solids; and hence the solids themselves are equal.

THEOREM XXXiv. (120.)

Every pyramid is a third part of a prism of equal base, and lying between the same parallel planes..

LET, first, the prism and pyramid have triangular bases; viz. the prism whose ends are BAC, EDF, and the pyramid whose base is DEF and vertex B: then the pyramid will be one-third part of the prism.