Of Solids 
bounded by 
P 
Fig. 148. 
oFig. 151, 
"Fig. 152, 
228 
9. A pyramid is a solid formed by several trian lax 
planes, which meet in a point: V, (ie. 148.) and ter- 
minate in the same plane rectilineal figure ABCDE, 
The plane figure ABCDE is called the base of the 
yramid; the point V-is its verter; and the triangles 
AVB, BVC, &c. taken together, form the convex or da- 
_ teral surface of the pyramid. 
10. The allitude of a pyramid, is the perpendicular 
drawn from its vertex to the plane-of its base, produced 
if necessary. 
11. A pyramid is triangular, nee ee &e. ac- 
cording as its base is a triangle, a qua ilateral, &c. 
12. A pyramid is. regular, when its base is a regular 
figure, and the perpendicular from its vertex passes 
through the centre of its base ; that is, through the cen- 
tre of a circle that may be-described aboutits base. 
13. Two solids are similar, when they are contained 
-by the same number of similar planes, similarly situa- 
ted, and having like inclinations to one another. 
“Prop. I. 
Two prisms are equal, when three planes, which con- 
‘tain a solid angle of the one, are equal to three planes 
which contain a solid angle of the other, each to each, 
and are similarly situated. 
‘THEOREM. 
Let the base ABCDE be equal .to the base abc de; 
the parallelogram ABGF equal to the parallelogram 
abgf, and the parallelogram BCHG equal to the pa- 
rallelogram 6c Ag; then shall the prism ABCDE- 
FGHIK be equal to the prism abcde-fg hik. 
For if the base ABCDE be placed upon its equal 
abcde, they will entirely coincide ; and because the 
three plane angles which form the solid angle B are 
equal to the three plane angles which form the solid 
angle b, each toeach, namely, ABC=a bc, ABG=abg, 
and GBC=g bc; and besides, these angles are similar- 
ly situated ; the solid angles B and 6 shall be equal 
(18.1.), and consequently the side. BG will fall on bg. 
Also, because of the equal parallelograms ABGF, ab g ify 
the side GF will fall on its equal gf, and, similarly, 
GH on gh: Therefore the upper base FGHIK will co- 
incide entirely with its equal fg hik, and the two so- 
lids will coincide entirely, and be equal to one another, 
Cor. Two right prisms which have equal bases and 
-equal altitudes are equal to one another. 
If the equal angles of the lower bases follow each 
ether in the same order, then the three planes which‘ 
contain each solid angle of the one prism will he respec- 
tively equal to three planes which contain a correspond- 
ing solid angle of the other, and will be similarly situated ; 
and when the one solid angle is applied to the other, 
these planes will coincide, and the prisms will exactly 
coincide. If the equal angles of the lower bases follow 
-each other in a contrary order, then, by inverting one 
of the prisms, so that its upper may become its lower 
base, the angles of the two bases will then be placed in 
the same order, so that in either case the prisms coin- 
cide, and are equal. 
Prop. IL. Tueror. 
In every parallelopiped, the opposite plan al 
and parallel. ini einai RH 
From the definition of this solid, the bases ABCD, 
EFGH are equal parallelograms ; and their sides are 
parallel: it remains therefore to demonstrate that the 
GEOMETRY. 
game is true for two opposite lateral faces, such as Of Sol 
AEHD, BFGC. Because the figure ABCD is:a paral- » 
lelogram, AD is equal and lel to BC; and for a 
like reason AE is equal and parallel to BF; therefore 
the angle DAE is equal to the angle CBF fanl) and 
the plane DAE is parallel to the plane CBF ; therefore 
also-the parallelogram DAEH is equal'to the: parallel. 
ogram CBFG. In like manner, it may. be demons ~ 
strated that the opposite parallelograms ABFE, DCGH . 
are equal and parallel. otf tuba ACY 
Cor. Any ‘two opposite faces of a’ pavallelopiped 
may be taken for its bases. _. ik} 
‘ 
Prop. If. Lemma, . 
In every prism ABCDE-FGHIK, ‘the ysections Fig. 1 
NOPQR, STVXY made by parallel planes are equal 
polygons. 
For the sides NO, ST are parallel, (7.1.) because 
‘the two parallel planes are cut by a third plane ABGF; 
these same sides NO, ST are comprehended between 
the parallels NS, OT, which are sides of the prism ; 
therefore NO=ST ; for a like reason the sides OP, PQ, 
‘QR, &c. of the section NOPQR are respectively equal 
to the sides TV, VX, XY, &c. of the section STVXY: 
Besides, these equal sides being at the same time pa- 
rallel, it follows that the angles NOP, OPQ, &c. of the 
first section are respectively equal to the angles STV, 
TVX, &c. of the second section. Therefore the two 
sections NOPQR, STVXY are equal polygons. © 
Cor. Every section of an upright prism by a plane 
parallel to the base is equal to that base. "pend UME 
Prop. IV. THrEor. 
If a parallelopiped AG, be eut by a plane’passing Fig. 154 
through BD, FH, the diagonals of two of the opposite 
Siakes it will be cut into two equivalent prisms, 
BAD-EFH, BCD-FGH. x pave} 
Through B and F, the extremities of one of the sides 
draw the planes Bade, F ehg perpendicular to 
to meet the three other sides of the solid in a, d, c, and 
ine, h, @; these sections are equal (3.), because the 
planes are perpendicular to FB, and therefore parallel. 
They are also parallelograms (7. 1.), because’ the op- 
posite sides of the same section, a B, dc are the inter- 
sections of two parallel planes ABFE, DCGH, by the 
same plane. ; ; ‘ 
For a like reason, the figure BaeF is a parallelo- 
gram, as also the other lateral faces BF gc, cdhg, 
a dhe, of the solid Badc-Fehg; therefore this so- 
lid is a prism (Def. 2.), and it is a right prism, because 
BF is perpendicular to the plane of its base... __ ‘oy 
This being premised, and it being observed that the 
right prism I 
prisms a Bd-e Fh, c Bd-gF kh, we shall now de- 
monstrate that the oblique triangular prism ABD- 
EFH is equal to the right triangular prism @Bd- 
e s ‘ : Be Ate these two prisms have a common part 
ABD-eF h, therefore it 1s onl y to 
that the remainders, viz. the solids BaAD L Fe Eun, f 
are equivalent to each other. Pet: 
Because BAEF, BaeF, are parallelograms, we have © 
AE=BF=ae, therefore Aa=Ee: In like manner, it 
may be proved that Dd=H: Conceive now that 
F ¢ h, the base of the solid F eEH A, is placed ona Bd, 
the base of the solid Ba AD d; then the point ¢ falling ~ 
his divided into two right triangular 
