GEOMETRY. 
ls on a, and hon d; the lines e E, h H will coincide with 
»>y their equals, a A, d D, because they are perpendiculars 
to the same plane: Therefore the two solids in ques- 
~ tion will coincide entirely, the one with the other, and 
hence it follows that the oblique prism BAD-FEH 
is equivalent to the right prism Bad-F e h. 
In the same manner it may be demonstrated, that 
the oblique prism BCD-FGH is equal to the right 
prism Bed-F gh; but the two right prisms are equal 
Cor. 1.), since they have the same altitude, and 
their bases are equal, they being halves of the same 
‘allelogram, ore the two triangular prisms 
BAD. EH, BCD-FGH which are equivalent to these 
"are equivalent to each other. 
Cor. Every triangular prism ABD-EFH is half a 
rallelopiped AG, having the same solid angle A, with 
me same edges AB, AD, AE, 
Scuouum. Although the triangular prisms into 
which the oblique parallelopiped is divided are con- 
tained by equal planes, and have their solid angles 
equal, yet they cannot be made to coincide. The rea- 
son is, that the plane angles about the corresponding 
solid angles in the two prisms are not placed in the 
same order. . These solid angles are therefore symmetri- 
cal, and cannot be brought to coincide. (18. 1.) Two 
prisms, or two solids of any kind so constituted, are 
called symmetrical solids. An exact notion of their re- 
lation to each other may be acquired by considering 
that any object and its image ected from a mirror 
are symmetrical figures. They resemble each other 
exactly, but every part is placed in a reverse order ; 
thus the reflected image of a right hand is a left hand. 
_ . In symmetrical solids, every circumstance upon which 
the magnitude of either depends, is the very same in, 
both, hence their equivalence might even be assumed as 
an axiom in solid geometry. 
Prop. V. Tneor. 
If two parallelopipeds, AG, AL, have a common base 
ABCD, and their upper bases EFGH, IKLM in the 
same plane, and between the same parallels EK, HL; 
these two parallelopipeds are equivalent to each other. 
There may be three cases, according as EI is greater 
or less than EF, or equal to it, but the demonstration 
is the same for them all. In the first place, the tri- 
angular prism AEI-DHM is. equal to the tri 
BFK-CGL; for since AE is parallel to BF, 
and EH to FG, the angle AEI= BFK, and HEI = 
GFK, and HEA=GFB;; of these six angles, the three 
first form the solid angle E, and the three others form the 
solid angle F ; therefore, since the plane angles are 
equal, each to each, and similarly situated, the solid 
Pe E and F are equal: and if the prism AEI- 
D be placed on the prism BFK-CGL, so that the 
base AEI may be on the base BFK, these being mani- 
festly equal, they will coincide ; and since the solid 
angle at E is equal to the solid angle at F, the side EH 
will fall on its equal FG, and coincide with it ; thus the 
planes which form the solid angles E and F will coin- 
cide, and the prisms will be equal (1.): Now if from the 
whole solid contained between the trapezoids AEKB, 
DHLC, there be taken the prism AEI-DHM, there 
will remain the llelopiped AIL, and if from the 
same solid there be taken the prism BFK-CGL, there 
will, remain the paralletoyioes AEG; therefore the 
oe ebrip AIL, AEG are equivalent to one 
3 
229 
Prop, VI. Tueor. 
Two bepalleloyipeds of the same base and the same 
altitude are equivalent to one another. 
Let ABCD be the common base of the two parallelo- Fig. 156. 
pipeds AG, AL; since they have the same altitude, 
their upper bases EFGH, IKLM will be in the same 
plane; also the sides EF, AB are equal and _ parallel, 
and the same is also true of IK and AB ; therefore EF 
is equal and parallel to IK; for a like reason GF is 
equal and parallel to LK. Let the sides EF and HG 
be produced, as also the sides LK and IM, so as to form 
by their intersection the parallelogram NOPQ; it is 
evident that this parallelogram is equal to each of the 
bases EFGH, IKLM. Now. if we suppose that there 
is a third parallelopiped, which, with the same lower 
base as the other two, has ‘for ‘its upper base NOPQ ; 
this third parallelopiped. will be equivalent to the pa- 
rallelopiped AG, (5,) and for a like reason it will be 
equivalent to the parallelopiped AL; therefore the two 
parallelopipeds AG, AL, which have the same base,. 
and the same altitude, are quivalent to one another. 
Prop. VII. 
Every peralalopiped is equivalent to a rectangular’ 
parallelopiped which has the same altitude and an equi- 
valent base. 
THEOR. 
Let AG be the proposed parallelopiped ; from the 
points A, B,C,D, draw AI, BK, CL, OM, perpendi- 
cular to the plane ABCD, and terminating in the plane 
of the upper base, and join IK, KL, LM, MI; thus 
there will be formed a par es AL equivalent to 
the parallelopiped AG, and of which the lateral faces AK, 
DL are rectangles. Ifthe base ABCD is also a rect- 
angle, AL will be a rectangular parallelopiped equiva- . 
lent to the proposed parallelopiped AG ; butif it is not, , 
(Fig. 157.) draw AO and BN perpendicular to CD, Fig. 157. 
and OQ and NP perpendiculars to the upper base, 
thus there will be formed a solid ABNO-IKPQ, 
which will be a rectangular parallelopiped. For, b 
construction, the base ABNO, and its opposite IKPt 
are rectangles, as also the lateral faces, because the edges 
AI, OQ are perpendicular to the plane of the base ; 
therefore the solid AP is a rectangular parallelopiped. 
But the two parallelopipeds AP, AL may be consider- 
ed as having the same base ABKI, and the same alti- 
tude AO ; therefore they are equivalent ; wherefore the 
parallelopiped AG, which was first transformed to the 
equivalent solid AL (Fig. 156.) is now reduced to the Fig. 156. 
eres ar AP Fe. 157.) Fig. 157. 
uivalent, rectangular 
which has the same height AI, anda base AONB equi- 
valent to the base ABCD. 
Prop. VIII. 
Two rectangular parallelopipeds AG, AL, which 
have the same ABCD, are to one another as their 
altitudes AE, Al. : 
Tueor. 
Let us suppose that AE contains some part of AI a 
certain number of times exactly, for example, let it con- 
tain the third part of Al five times, and let these equal 
parts be A p,pq, q1, Ir, r F: Let planes be supposed 
to pass through p,q, r parallel to the common base ; 
these will divide the solid AG into five parallelopipeds, 
bounded by 
Planes. 
Fig. 156. 
Fig. 158, 
