BOOK XI.
DEFINITIONS.
1
A
solid is that which has length, breadth, and depth.
2
An extremity of a solid is a surface.
3
A
straight line is
at right angles to a plane, when it makes right angles with all the straight lines which meet it and are in the plane.
4
A
plane is at right angles to a plane when the straight lines drawn, in one of the planes, at right angles to the common section of the planes are at right angles to the remaining plane.
5
The
inclination of a straight line to a plane is, assuming a perpendicular drawn from the extremity of the straight line which is elevated above the plane to the plane, and a straight line joined from the point thus arising to the extremity of the straight line which is in the plane, the angle contained by the straight line so drawn and the straight line standing up.
6
The
inclination of a plane to a plane is the acute angle contained by the straight lines drawn at right angles to the common section at the same point, one in each of the planes.
7
A plane is said to be
similarly inclined to a plane as another is to another when the said angles of the inclinations are equal to one another.
8
Parallel planes are those which do not meet.
9
Similar solid figures are those contained by similar planes equal in multitude.
10
Equal and similar solid figures are those contained by similar planes equal in multitude and in magnitude.
11
A
solid angle is the inclination constituted by more than two lines which meet one another and are not in the same surface, towards all the lines.
Otherwise: A
solid angle is that which is contained by more than two plane angles which are not in the same plane and are constructed to one point.
12
A
pyramid is a solid figure, contained by planes, which is constructed from one plane to one point.
13
A
prism is a solid figure contained by planes two of which, namely those which are opposite, are equal, similar and parallel, while the rest are parallelograms.
14
When, the diameter of a semicircle remaining fixed, the semicircle is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a
sphere.
15
The
axis of the sphere is the straight line which remains fixed and about which the semicircle is turned.
16
The
centre of the sphere is the same as that of the semicircle.
17
A
diameter of the sphere is any straight line drawn through the centre and terminated in both directions by the surface of the sphere.
18
When, one side of those about the right angle in a right-angled triangle remaining fixed, the triangle is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a
cone.
And, if the straight line which remains fixed be equal to the remaining side about the right angle which is carried round, the cone will be
right-angled; if less,
obtuse-angled; and if greater,
acute-angled.
19
The
axis of the cone is the straight line which remains fixed and about which the triangle is turned.
20
And the
base is the circle described by the straight line which is carried round.
21
When, one side of those about the right angle in a rectangular parallelogram remaining fixed, the parallelogram is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a
cylinder.
22
The
axis of the cylinder is the straight line which remains fixed and about which the parallelogram is turned.
23
And the
bases are the circles described by the two sides opposite to one another which are carried round.
24
Similar cones and cylinders are those in which the axes and the diameters of the bases are proportional.
25
A
cube is a solid figure contained by six equal squares.
26
An
octahedron is a solid figure contained by eight equal and equilateral triangles.
27
An
icosahedron is a solid figure contained by twenty equal and equilateral triangles.
28
A
dodecahedron is a solid figure contained by twelve equal, equilateral, and equiangular pentagons.
BOOK XI. PROPOSITIONS.
PROPOSITION 1.
A part of a straight line cannot be in the plane of reference and a part in a plane more elevated.
For, if possible, let a part
AB of the straight line
ABC be in the plane of reference, and a part
BC in a plane more elevated.
There will then be in the plane of reference some straight line continuous with
AB in a straight line.
Let it be
BD; therefore
AB is a common segment of the two straight lines
ABC,
ABD: which is impossible, inasmuch as, if we describe a circle with centre
B and distance
AB, the diameters will cut off unequal circumferences of the circle.
Therefore a part of a straight line cannot be in the plane of reference, and a part in a plane more elevated. Q. E. D.
1
2
PROPOSITION 2.
If two straight lines cut one another,
they are in one plane,
and every triangle is in one plane.
For let the two straight lines
AB,
CD cut one another at the point
E; I say that
AB,
CD are in one plane, and every triangle is in one plane.
For let points
F,
G be taken at random on
EC,
EB, let
CB,
FG be joined, and let
FH,
GK be drawn across; I say first that the triangle
ECB is in one plane.
For, if part of the triangle
ECB, either
FHC or
GBK, is in the plane of reference, and the rest in another, a part also of one of the straight lines
EC,
EB will be in the plane of reference, and a part in another.
But, if the part
FCBG of the triangle
ECB be in the plane of reference, and the rest in another, a part also of both the straight lines
EC,
EB will be in the plane of reference and a part in another: which was proved absurd. [
XI. 1]
Therefore the triangle
ECB is in one plane.
But, in whatever plane the triangle
ECB is, in that plane also is each of the straight lines
EC,
EB, and, in whatever plane each of the straight lines
EC,
EB is, in that plane are
AB,
CD also. [
XI. 1]
Therefore the straight lines
AB,
CD are in one plane, and every triangle is in one plane. Q. E. D.
PROPOSITION 3.
If two planes cut one another,
their common section is a straight line.
For let the two planes
AB,
BC cut one another, and let the line
DB be their common section; I say that the line
DB is a straight line.
For, if not, from
D to
B let the straight line
DEB be joined in the plane
AB, and in the plane
BC the straight line
DFB.
Then the two straight lines
DEB,
DFB will have the same extremities, and will clearly enclose an area: which is absurd.
Therefore
DEB,
DFB are not straight lines.
Similarly we can prove that neither will there be any other straight line joined from
D to
B except
DB the common section of the planes
AB,
BC.
Therefore etc. Q. E. D.
PROPOSITION 4.
If a straight line be set up at right angles to two straight lines which cut one another,
at their common point of section,
it will also be at right angles to the plane through them.
For let a straight line
EF be set up at right angles to the two straight lines
AB,
CD, which cut one another at the point
E, from
E; I say that
EF is also at right angles to the plane through
AB,
CD.
For let
AE,
EB,
CE,
ED be cut off equal to one another, and let any straight line
GEH be drawn across through
E, at random; let
AD,
CB be joined, and further let
FA,
FG,
FD,
FC,
FH,
FB be joined from the point
F taken at random <on
EF>.
Now, since the two straight lines
AE,
ED are equal to the two straight lines
CE,
EB, and contain equal angles, [
I. 15] therefore the base
AD is equal to the base
CB, and the triangle
AED will be equal to the triangle
CEB; [
I. 4] so that the angle
DAE is also equal to the angle
EBC.
But the angle
AEG is also equal to the angle
BEH; [
I. 15] therefore
AGE,
BEH are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely that adjacent to the equal angles, that is to say,
AE to
EB; therefore they will also have the remaining sides equal to the remaining sides. [
I. 26]
Therefore
GE is equal to
EH, and
AG to
BH.
And, since
AE is equal to
EB, while
FE is common and at right angles, therefore the base
FA is equal to the base
FB. [
I. 4]
For the same reason
FC is also equal to
FD.
And, since
AD is equal to
CB, and
FA is also equal to
FB, the two sides
FA,
AD are equal to the two sides
FB,
BC respectively; and the base
FD was proved equal to the base
FC; therefore the angle
FAD is also equal to the angle
FBC. [
I. 8]
And since, again,
AG was proved equal to
BH, and further
FA also equal to
FB, the two sides
FA,
AG are equal to the two sides
FB,
BH.
And the angle
FAG was proved equal to the angle
FBH; therefore the base
FG is equal to the base
FH. [
I. 4]
Now since, again,
GE was proved equal to
EH, and
EF is common, the two sides
GE,
EF are equal to the two sides
HE,
EF; and the base
FG is equal to the base
FH; therefore the angle
GEF is equal to the angle
HEF. [
I. 8]
Therefore each of the angles
GEF,
HEF is right.
Therefore
FE is at right angles to
GH drawn at random through
E.
Similarly we can prove that
FE will also make right angles with all the straight lines which meet it and are in the plane of reference.
But a straight line is at right angles to a plane when it makes right angles with all the straight lines which meet it and are in that same plane; [
XI. Def. 3] therefore
FE is at right angles to the plane of reference.
But the plane of reference is the plane through the straight lines
AB,
CD.
Therefore
FE is at right angles to the plane through
AB,
CD.
Therefore etc. Q. E. D.
PROPOSITION 5.
If a straight line be set up at right angles to three straight lines which meet one another,
at their common point of section,
the three straight lines are in one plane.
For let a straight line
AB be set up at right angles to the three straight lines
BC,
BD,
BE, at their point of meeting at
B; I say that
BC,
BD,
BE are in one plane.
For suppose they are not, but, if possible, let
BD,
BE be in the plane of reference and
BC in one more elevated; let the plane through
AB,
BC be produced; it will thus make, as common section in the plane of reference, a straight line. [
XI. 3]
Let it make
BF.
Therefore the three straight lines
AB,
BC,
BF are in one plane, namely that drawn through
AB,
BC.
Now, since
AB is at right angles to each of the straight lines
BD,
BE, therefore
AB is also at right angles to the plane through
BD,
BE. [
XI. 4]
But the plane through
BD,
BE is the plane of reference; therefore
AB is at right angles to the plane of reference.
Thus
AB will also make right angles with all the straight lines which meet it and are in the plane of reference. [
XI. Def. 3]
But
BF which is in the plane of reference meets it; therefore the angle
ABF is right.
But, by hypothesis, the angle
ABC is also right; therefore the angle
ABF is equal to the angle
ABC.
And they are in one plane: which is impossible.
Therefore the straight line
BC is not in a more elevated plane; therefore the three straight lines
BC,
BD,
BE are in one plane.
Therefore, if a straight line be set up at right angles to three straight lines, at their point of meeting, the three straight lines are in one plane. Q. E. D.
PROPOSITION 6.
If two straight lines be at right angles to the same plane,
the straight lines will be parallel.
For let the two straight lines
AB,
CD be at right angles to the plane of reference; I say that
AB is parallel to
CD.
For let them meet the plane of reference at the points
B,
D, let the straight line
BD be joined, let
DE be drawn, in the plane of reference, at right angles to
BD, let
DE be made equal to
AB, and let
BE,
AE,
AD be joined.
Now, since
AB is at right angles to the plane of reference, it will also make right angles with all the straight lines which meet it and are in the plane of reference. [
XI. Def. 3]
But each of the straight lines
BD,
BE is in the plane of reference and meets
AB; therefore each of the angles
ABD,
ABE is right.
For the same reason each of the angles
CDB,
CDE is also right.
And, since
AB is equal to
DE, and
BD is common, the two sides
AB,
BD are equal to the two sides
ED,
DB; and they include right angles; therefore the base
AD is equal to the base
BE. [
I. 4]
And, since
AB is equal to
DE, while
AD is also equal to
BE, the two sides
AB,
BE are equal to the two sides
ED,
DA; and
AE is their common base; therefore the angle
ABE is equal to the angle
EDA. [
I. 8]
But the angle
ABE is right; therefore the angle
EDA is also right; therefore
ED is at right angles to
DA.
But it is also at right angles to each of the straight lines
BD,
DC; therefore
ED is set up at right angles to the three straight lines
BD,
DA,
DC at their point of meeting; therefore the three straight lines
BD,
DA,
DC are in one plane. [
XI. 5]
But, in whatever plane
DB,
DA are, in that plane is
AB also, for every triangle is in one plane; [
XI. 2] therefore the straight lines
AB,
BD,
DC are in one plane.
And each of the angles
ABD,
BDC is right; therefore
AB is parallel to
CD. [
I. 28]
Therefore etc. Q. E. D.
PROPOSITION 7.
If two straight lines be parallel and points be taken at random on each of them,
the straight line joining the points is in the same plane with the parallel straight lines.
Let
AB,
CD be two parallel straight lines, and let points
E,
F be taken at random on them respectively; I say that the straight line joining the points
E,
F is in the same plane with the parallel straight lines.
For suppose it is not, but, if possible, let it be in a more elevated plane as
EGF, and let a plane be drawn through
EGF; it will then make, as section in the plane of reference, a straight line. [
XI. 3]
Let it make it, as
EF; therefore the two straight lines
EGF,
EF will enclose an area: which is impossible.
Therefore the straight line joined from
E to
F is not in a plane more elevated; therefore the straight line joined from
E to
F is in the plane through the parallel straight lines
AB,
CD.
Therefore etc. Q. E. D.
PROPOSITION 8.
If two straight lines be parallel,
and one of them be at right angles to any plane,
the remaining one will also be at right angles to the same plane.
Let
AB,
CD be two parallel straight lines, and let one of them,
AB, be at right angles to the plane of reference; I say that the remaining one,
CD, will also be at right angles to the same plane.
For let
AB,
CD meet the plane of reference at the points
B,
D, and let
BD be joined; therefore
AB,
CD,
BD are in one plane. [
XI. 7]
Let
DE be drawn, in the plane of reference, at right angles to
BD, let
DE be made equal to
AB, and let
BE,
AE,
AD be joined.
Now, since
AB is at right angles to the plane of reference, therefore
AB is also at right angles to all the straight lines which meet it and are in the plane of reference; [
XI. Def. 3] therefore each of the angles
ABD,
ABE is right.
And, since the straight line
BD has fallen on the parallels
AB,
CD, therefore the angles
ABD,
CDB are equal to two right angles. [
I. 29]
But the angle
ABD is right; therefore the angle
CDB is also right; therefore
CD is at right angles to
BD.
And, since
AB is equal to
DE, and
BD is common, the two sides
AB,
BD are equal to the two sides
ED,
DB; and the angle
ABD is equal to the angle
EDB, for each is right; therefore the base
AD is equal to the base
BE.
And, since
AB is equal to
DE, and
BE to
AD, the two sides
AB,
BE are equal to the two sides
ED,
DA respectively, and
AE is their common base; therefore the angle
ABE is equal to the angle
EDA.
But the angle
ABE is right; therefore the angle
EDA is also right; therefore
ED is at right angles to
AD.
But it is also at right angles to
DB; therefore
ED is also at right angles to the plane through
BD,
DA. [
XI. 4]
Therefore
ED will also make right angles with all the straight lines which meet it and are in the plane through
BD,
DA.
But
DC is in the plane through
BD,
DA, inasmuch as
AB,
BD are in the plane through
BD,
DA, [
XI. 2] and
DC is also in the plane in which
AB,
BD are.
Therefore
ED is at right angles to
DC, so that
CD is also at right angles to
DE.
But
CD is also at right angles to
BD.
Therefore
CD is set up at right angles to the two straight lines
DE,
DB which cut one another, from the point of section at
D; so that
CD is also at right angles to the plane through
DE,
DB. [
XI. 4]
But the plane through
DE,
DB is the plane of reference; therefore
CD is at right angles to the plane of reference.
Therefore etc. Q. E. D.
PROPOSITION 9.
Straight lines which are parallel to the same straight line and are not in the same plane with it are also parallel to one another.
For let each of the straight lines
AB,
CD be parallel to
EF, not being in the same plane with it; I say that
AB is parallel to
CD.
For let a point
G be taken at random on
EF, and from it let there be drawn
GH, in the plane through
EF,
AB, at right angles to
EF, and
GK in the plane through
FE,
CD again at right angles to
EF.
Now, since
EF is at right angles to each of the straight lines
GH,
GK, therefore
EF is also at right angles to the plane through
GH,
GK. [
XI. 4]
And
EF is parallel to
AB; therefore
AB is also at right angles to the plane through
HG,
GK. [
XI. 8]
For the same reason
CD is also at right angles to the plane through
HG,
GK; therefore each of the straight lines
AB,
CD is at right angles to the plane through
HG,
GK.
But if two straight lines be at right angles to the same plane, the straight lines are parallel; [
XI. 6] therefore
AB is parallel to
CD.
PROPOSITION 10.
If two straight lines meeting one another be parallel to two straight lines meeting one another not in the same plane,
they will contain equal angles.
For let the two straight lines
AB,
BC meeting one another be parallel to the two straight lines
DE,
EF meeting one another, not in the same plane; I say that the angle
ABC is equal to the angle
DEF.
For let
BA,
BC,
ED,
EF be cut off equal to one another, and let
AD,
CF,
BE,
AC,
DF be joined.
Now, since
BA is equal and parallel to
ED, therefore
AD is also equal and parallel to
BE. [
I. 33]
For the same reason
CF is also equal and parallel to
BE.
Therefore each of the straight lines
AD,
CF is equal and parallel to
BE.
But straight lines which are parallel to the same straight line and are not in the same plane with it are parallel to one another; [
XI. 9] therefore
AD is parallel and equal to
CF.
And
AC,
DF join them; therefore
AC is also equal and parallel to
DF. [
I. 33]
Now, since the two sides
AB,
BC are equal to the two sides
DE,
EF, and the base
AC is equal to the base
DF, therefore the angle
ABC is equal to the angle
DEF. [
I. 8]
Therefore etc.
PROPOSITION 11.
From a given elevated point to draw a straight line perpendicular to a given plane.
Let
A be the given elevated point, and the plane of reference the given plane; thus it is required to draw from the point
A a straight line perpendicular to the plane of reference.
Let any straight line
BC be drawn, at random, in the plane of reference, and let
AD be drawn from the point
A perpendicular to
BC. [
I. 12]
If then
AD is also perpendicular to the plane of reference, that which was enjoined will have been done.
But, if not, let
DE be drawn from the point
D at right angles to
BC and in the plane of reference, [
I. 11] let
AF be drawn from
A perpendicular to
DE, [
I. 12] and let
GH be drawn through the point
F parallel to
BC. [
I. 31]
Now, since
BC is at right angles to each of the straight lines
DA,
DE, therefore
BC is also at right angles to the plane through
ED,
DA. [
XI. 4]
And
GH is parallel to it; but, if two straight lines be parallel, and one of them be at right angles to any plane, the remaining one will also be at right angles to the same plane; [
XI. 8] therefore
GH is also at right angles to the plane through
ED,
DA.
Therefore
GH is also at right angles to all the straight lines which meet it and are in the plane through
ED,
DA. [
XI. Def. 3]
But
AF meets it and is in the plane through
ED,
DA; therefore
GH is at right angles to
FA, so that
FA is also at right angles to
GH.
But
AF is also at right angles to
DE; therefore
AF is at right angles to each of the straight lines
GH,
DE.
But, if a straight line be set up at right angles to two straight lines which cut one another, at the point of section, it will also be at right angles to the plane through them; [
XI. 4] therefore
FA is at right angles to the plane through
ED,
GH.
But the plane through
ED,
GH is the plane of reference; therefore
AF is at right angles to the plane of reference.
Therefore from the given elevated point
A the straight line
AF has been drawn perpendicular to the plane of reference. Q. E. F.
PROPOSITION 12.
To set up a straight line at right angles to a given plane from a given point in it.
Let the plane of reference be the given plane, and
A the point in it; thus it is required to set up from the point
A a straight line at right angles to the plane of reference.
Let any elevated point
B be conceived, from
B let
BC be drawn perpendicular to the plane of reference, [
XI. 11] and through the point
A let
AD be drawn parallel to
BC. [
I. 31]
Then, since
AD,
CB are two parallel straight lines, while one of them,
BC, is at right angles to the plane of reference, therefore the remaining one,
AD, is also at right angles to the plane of reference. [
XI. 8]
Therefore
AD has been set up at right angles to the given plane from the point
A in it.
PROPOSITION 13.
From the same point two straight lines cannot be set up at right angles to the same plane on the same side.
For, if possible, from the same point
A let the two straight lines
AB,
AC be set up at right angles to the plane of reference and on the same side, and let a plane be drawn through
BA,
AC; it will then make, as section through
A in the plane of reference, a straight line. [
XI. 3]
Let it make
DAE; therefore the straight lines
AB,
AC,
DAE are in one plane.
And, since
CA is at right angles to the plane of reference, it will also make right angles with all the straight lines which meet it and are in the plane of reference. [
XI. Def. 3]
But
DAE meets it and is in the plane of reference; therefore the angle
CAE is right.
For the same reason the angle
BAE is also right; therefore the angle
CAE is equal to the angle
BAE.
And they are in one plane: which is impossible.
Therefore etc. Q. E. D.
PROPOSITION 14.
Planes to which the same straight line is at right angles will be parallel.
For let any straight line
AB be at right angles to each of the planes
CD,
EF; I say that the planes are parallel.
For, if not, they will meet when produced.
Let them meet; they will then make, as common section, a straight line. [
XI. 3]
Let them make
GH; let a point
K be taken at random on
GH, and let
AK,
BK be joined.
Now, since
AB is at right angles to the plane
EF, therefore
AB is also at right angles to
BK which is a straight line in the plane
EF produced; [
XI. Def. 3] therefore the angle
ABK is right.
For the same reason the angle
BAK is also right.
Thus, in the triangle
ABK, the two angles
ABK,
BAK are equal to two right angles: which is impossible. [
I. 17]
Therefore the planes
CD,
EF will not meet when produced; therefore the planes
CD,
EF are parallel. [
XI. Def. 8]
Therefore planes to which the same straight line is at right angles are parallel. Q. E. D.
PROPOSITION 15.
If two straight lines meeting one another be parallel to two straight lines meeting one another,
not being in the same plane,
the planes through them are parallel.
For let the two straight lines
AB,
BC meeting one another be parallel to the two straight lines
DE,
EF meeting one another, not being in the same plane; I say that the planes produced through
AB,
BC and
DE,
EF will not meet one another.
For let
BG be drawn from the point
B perpendicular to the plane through
DE,
EF [
XI. 11], and let it meet the plane at the point
G; through
G let
GH be drawn parallel to
ED, and
GK parallel to
EF. [
I. 31]
Now, since
BG is at right angles to the plane through
DE,
EF, therefore it will also make right angles with all the straight lines which meet it and are in the plane through
DE,
EF. [
XI. Def. 3]
But each of the straight lines
GH,
GK meets it and is in the plane through
DE,
EF; therefore each of the angles
BGH,
BGK is right.
And, since
BA is parallel to
GH, [
XI. 9] therefore the angles
GBA,
BGH are equal to two right angles. [
I. 29]
But the angle
BGH is right; therefore the angle
GBA is also right; therefore
GB is at right angles to
BA.
For the same reason
GB is also at right angles to
BC.
Since then the straight line
GB is set up at right angles to the two straight lines
BA,
BC which cut one another, therefore
GB is also at right angles to the plane through
BA,
BC. [
XI. 4]
But planes to which the same straight line is at right angles are parallel; [
XI. 14] therefore the plane through
AB,
BC is parallel to the plane through
DE,
EF.
Therefore, if two straight lines meeting one another be parallel to two straight lines meeting one another, not in the same plane, the planes through them are parallel. Q. E. D.
PROPOSITION 16.
If two parallel planes be cut by any plane,
their common sections are parallel.
For let the two parallel planes
AB,
CD be cut by the plane
EFGH, and let
EF,
GH be their common sections; I say that
EF is parallel to
GH.
For, if not,
EF,
GH will, when produced, meet either in the direction of
F,
H or of
E,
G.
Let them be produced, as in the direction of
F,
H, and let them, first, meet at
K.
Now, since
EFK is in the plane
AB, therefore all the points on
EFK are also in the plane
AB. [
XI. 1]
But
K is one of the points on the straight line
EFK; therefore
K is in the plane
AB.
For the same reason
K is also in the plane
CD; therefore the planes
AB,
CD will meet when produced.
But they do not meet, because they are, by hypothesis, parallel; therefore the straight lines
EF,
GH will not meet when produced in the direction of
F,
H.
Similarly we can prove that neither will the straight lines
EF,
GH meet when produced in the direction of
E,
G.
But straight lines which do not meet in either direction are parallel. [
I. Def. 23]
Therefore
EF is parallel to
GH.
Therefore etc. Q. E. D.
PROPOSITION 17.
If two straight lines be cut by parallel planes,
they will be cut in the same ratios.
For let the two straight lines
AB,
CD be cut by the parallel planes
GH,
KL,
MN at the points
A,
E,
B and
C,
F,
D; I say that, as the straight line
AE is to
EB, so is
CF to
FD.
For let
AC,
BD,
AD be joined, let
AD meet the plane
KL at the point
O, and let
EO,
OF be joined.
Now, since the two parallel planes
KL,
MN are cut by the plane
EBDO, their common sections
EO,
BD are parallel. [
XI. 16]
For the same reason, since the two parallel planes
GH,
KL are cut by the plane
AOFC, their common sections
AC,
OF are parallel. [
id.]
And, since the straight line
EO has been drawn parallel to
BD, one of the sides of the triangle
ABD, therefore, proportionally, as
AE is to
EB, so is
AO to
OD. [
VI. 2]
Again, since the straight line
OF has been drawn parallel to
AC, one of the sides of the triangle
ADC, proportionally, as
AO is to
OD, so is
CF to
FD. [
id.]
But it was also proved that, as
AO is to
OD, so is
AE to
EB; therefore also, as
AE is to
EB, so is
CF to
FD. [
V. 11]
Therefore etc. Q. E. D.
PROPOSITION 18.
If a straight line be at right angles to any plane,
all the planes through it will also be at right angles to the same plane.
For let any straight line
AB be at right angles to the plane of reference; I say that all the planes through
AB are also at right angles to the plane of reference.
For let the plane
DE be drawn through
AB, let
CE be the common section of the plane
DE and the plane of reference, let a point
F be taken at random on
CE, and from
F let
FG be drawn in the plane
DE at right angles to
CE. [
I. 11]
Now, since
AB is at right angles to the plane of reference,
AB is also at right angles to all the straight lines which meet it and are in the plane of reference; [
XI. Def. 3] so that it is also at right angles to
CE; therefore the angle
ABF is right.
But the angle
GFB is also right; therefore
AB is parallel to
FG. [
I. 28]
But
AB is at right angles to the plane of reference; therefore
FG is also at right angles to the plane of reference. [
XI. 8]
Now a plane is at right angles to a plane, when the straight lines drawn, in one of the planes, at right angles to the common section of the planes are at right angles to the remaining plane. [
XI. Def. 4]
And
FG, drawn in one of the planes
DE at right angles to
CE, the common section of the planes, was proved to be at right angles to the plane of reference; therefore the plane
DE is at right angles to the plane of reference.
Similarly also it can be proved that all the planes through
AB are at right angles to the plane of reference.
Therefore etc. Q. E. D.
PROPOSITION 19.
If two planes which cut one another be at right angles to any plane,
their common section will also be at right angles to the same plane.
For let the two planes
AB,
BC be at right angles to the plane of reference, and let
BD be their common section; I say that
BD is at right angles to the plane of reference.
For suppose it is not, and from the point
D let
DE be drawn in the plane
AB at right angles to the straight line
AD, and
DF in the plane
BC at right angles to
CD.
Now, since the plane
AB is at right angles to the plane of reference, and
DE has been drawn in the plane
AB at right angles to
AD, their common section, therefore
DE is at right angles to the plane of reference. [
XI. Def. 4]
Similarly we can prove that
DF is also at right angles to the plane of reference.
Therefore from the same point
D two straight lines have been set up at right angles to the plane of reference on the same side: which is impossible. [
XI. 13]
Therefore no straight line except the common section
DB of the planes
AB,
BC can be set up from the point
D at right angles to the plane of reference.
Therefore etc. Q. E. D.
PROPOSITION 20.
If a solid angle be contained by three plane angles,
any two,
taken together in any manner,
are greater than the remaining one.
For let the solid angle at
A be contained by the three plane angles
BAC,
CAD,
DAB; I say that any two of the angles
BAC,
CAD,
DAB, taken together in any manner, are greater than the remaining one.
If now the angles
BAC,
CAD,
DAB are equal to one another, it is manifest that any two are greater than the remaining one.
But, if not, let
BAC be greater, and on the straight line
AB, and at the point
A on it, let the angle
BAE be constructed, in the plane through
BA,
AC, equal to the angle
DAB; let
AE be made equal to
AD, and let
BEC, drawn across through the point
E, cut the straight lines
AB,
AC at the points
B,
C; let
DB,
DC be joined.
Now, since
DA is equal to
AE, and
AB is common, two sides are equal to two sides; and the angle
DAB is equal to the angle
BAE; therefore the base
DB is equal to the base
BE. [
I. 4]
And, since the two sides
BD,
DC are greater than
BC, [
I. 20] and of these
DB was proved equal to
BE, therefore the remainder
DC is greater than the remainder
EC.
Now, since
DA is equal to
AE, and
AC is common, and the base
DC is greater than the base
EC, therefore the angle
DAC is greater than the angle
EAC. [
I. 25]
But the angle
DAB was also proved equal to the angle
BAE; therefore the angles
DAB,
DAC are greater than the angle
BAC.
Similarly we can prove that the remaining angles also, taken together two and two, are greater than the remaining one.
Therefore etc. Q. E. D.
PROPOSITION 21.
Any solid angle is contained by plane angles less than four right angles.
Let the angle at
A be a solid angle contained by the plane angles
BAC,
CAD,
DAB; I say that the angles
BAC,
CAD,
DAB are less than four right angles.
For let points
B,
C,
D be taken at random on the straight lines
AB,
AC,
AD respectively, and let
BC,
CD,
DB be joined.
Now, since the solid angle at
B is contained by the three plane angles
CBA,
ABD,
CBD, any two are greater than the remaining one; [
XI. 20] therefore the angles
CBA,
ABD are greater than the angle
CBD.
For the same reason the angles
BCA,
ACD are also greater than the angle
BCD, and the angles
CDA,
ADB are greater than the angle
CDB; therefore the six angles
CBA,
ABD,
BCA,
ACD,
CDA,
ADB are greater than the three angles
CBD,
BCD,
CDB.
But the three angles
CBD,
BDC,
BCD are equal to two right angles; [
I. 32] therefore the six angles
CBA,
ABD,
BCA,
ACD,
CDA,
ADB are greater than two right angles.
And, since the three angles of each of the triangles
ABC,
ACD,
ADB are equal to two right angles, therefore the nine angles of the three triangles, the angles
CBA,
ACB,
BAC,
ACD,
CDA,
CAD,
ADB,
DBA,
BAD are equal to six right angles; and of them the six angles
ABC,
BCA,
ACD,
CDA,
ADB,
DBA are greater than two right angles; therefore the remaining three angles
BAC,
CAD,
DAB containing the solid angle are less than four right angles.
Therefore etc. Q. E. D.
PROPOSITION 22.
If there be three plane angles of which two,
taken together in any manner,
are greater than the remaining one,
and they are contained by equal straight lines,
it is possible to construct a triangle out of the straight lines joining the extremities of the equal straight lines.
Let there be three plane angles
ABC,
DEF,
GHK, of which two, taken together in any manner, are greater than the remaining one, namely
the angles ABC, DEF greater than the angle GHK, the angles DEF, GHK greater than the angle ABC, and, further, the angles
GHK,
ABC greater than the angle
DEF; let the straight lines
AB,
BC,
DE,
EF,
GH,
HK be equal, and let
AC,
DF,
GK be joined; I say that it is possible to construct a triangle out of straight lines equal to
AC,
DF,
GK, that is, that any two of the straight lines
AC,
DF,
GK are greater than the remaining one.
Now, if the angles
ABC,
DEF,
GHK are equal to one another, it is manifest that,
AC,
DF,
GK being equal also, it is possible to construct a triangle out of straight lines equal to
AC,
DF,
GK.
But, if not, let them be unequal, and on the straight line
HK, and at the point
H on it, let the angle
KHL be constructed equal to the angle
ABC; let
HL be made equal to one of the straight lines
AB,
BC,
DE,
EF,
GH,
HK, and let
KL,
GL be joined.
Now, since the two sides
AB,
BC are equal to the two sides
KH,
HL, and the angle at
B is equal to the angle
KHL, therefore the base
AC is equal to the base
KL. [
I. 4]
And, since the angles
ABC,
GHK are greater than the angle
DEF, while the angle
ABC is equal to the angle
KHL, therefore the angle
GHL is greater than the angle
DEF.
And, since the two sides
GH,
HL are equal to the two sides
DE,
EF, and the angle
GHL is greater than the angle
DEF, therefore the base
GL is greater than the base
DF. [
I. 24]
But
GK,
KL are greater than
GL.
Therefore
GK,
KL are much greater than
DF.
But
KL is equal to
AC; therefore
AC,
GK are greater than the remaining straight line
DF.
Similarly we can prove that
AC,
DF are greater than
GK, and further
DF,
GK are greater than
AC.
Therefore it is possible to construct a triangle out of straight lines equal to
AC,
DF,
GK. Q. E. D.
PROPOSITION 23.
To construct a solid angle out of three plane angles two of which,
taken together in any manner,
are greater than the remaining one: thus the three angles must be less than four right angles.
Let the angles
ABC,
DEF,
GHK be the three given plane angles, and let two of these, taken together in any manner, be greater than the remaining one, while, further, the three are less than four right angles; thus it is required to construct a solid angle out of angles equal to the angles
ABC,
DEF,
GHK.
Let
AB,
BC,
DE,
EF,
GH,
HK be cut off equal to one another, and let
AC,
DF,
GK be joined; it is therefore possible to construct a triangle out of straight lines equal to
AC,
DF,
GK. [
XI. 22]
Let
LMN be so constructed that
AC is equal to
LM,
DF to
MN, and further
GK to
NL, let the circle
LMN be described about the triangle
LMN, let its centre be taken, and let it be
O; let
LO,
MO,
NO be joined; I say that
AB is greater than
LO.
For, if not,
AB is either equal to
LO, or less.
First, let it be equal.
Then, since
AB is equal to
LO, while
AB is equal to
BC, and
OL to
OM, the two sides
AB,
BC are equal to the two sides
LO,
OM respectively; and, by hypothesis, the base
AC is equal to the base
LM; therefore the angle
ABC is equal to the angle
LOM. [
I. 8]
For the same reason the angle
DEF is also equal to the angle
MON, and further the angle
GHK to the angle
NOL; therefore the three angles
ABC,
DEF,
GHK are equal to the three angles
LOM,
MON,
NOL.
But the three angles
LOM,
MON,
NOL are equal to four right angles; therefore the angles
ABC,
DEF,
GHK are equal to four right angles.
But they are also, by hypothesis, less than four right angles: which is absurd.
Therefore
AB is not equal to
LO.
I say next that neither is
AB less than
LO.
For, if possible, let it be so, and let
OP be made equal to
AB, and
OQ equal to
BC, and let
PQ be joined.
Then, since
AB is equal to
BC,
OP is also equal to
OQ, so that the remainder
LP is equal to
QM.
Therefore
LM is parallel to
PQ, [
VI. 2] and
LMO is equiangular with
PQO; [
I. 29] therefore, as
OL is to
LM, so is
OP to
PQ; [
VI. 4] and alternately, as
LO is to
OP, so is
LM to
PQ. [
V. 16]
But
LO is greater than
OP; therefore
LM is also greater than
PQ.
But
LM was made equal to
AC; therefore
AC is also greater than
PQ.
Since, then, the two sides
AB,
BC are equal to the two sides
PO,
OQ, and the base
AC is greater than the base
PQ, therefore the angle
ABC is greater than the angle
POQ. [
I. 25]
Similarly we can prove that the angle
DEF is also greater than the angle
MON, and the angle
GHK greater than the angle
NOL.
Therefore the three angles
ABC,
DEF,
GHK are greater than the three angles
LOM,
MON,
NOL.
But, by hypothesis, the angles
ABC,
DEF,
GHK are less than four right angles; therefore the angles
LOM,
MON,
NOL are much less than four right angles.
But they are also equal to four right angles: which is absurd.
Therefore
AB is not less than
LO.
And it was proved that neither is it equal; therefore
AB is greater than
LO.
Let then
OR be set up from the point
O at right angles to the plane of the circle
LMN, [
XI. 12] and let the square on
OR be equal to that area by which the square on
AB is greater than the square on
LO; [Lemma] let
RL,
RM,
RN be joined.
Then, since
RO is at right angles to the plane of the circle
LMN, therefore
RO is also at right angles to each of the straight lines
LO,
MO,
NO.
And, since
LO is equal to
OM, while
OR is common and at right angles, therefore the base
RL is equal to the base
RM. [
I. 4]
For the same reason
RN is also equal to each of the straight lines
RL,
RM; therefore the three straight lines
RL,
RM,
RN are equal to one another.
Next, since by hypothesis the square on
OR is equal to that area by which the square on
AB is greater than the square on
LO, therefore the square on
AB is equal to the squares on
LO,
OR.
But the square on
LR is equal to the squares on
LO,
OR, for the angle
LOR is right; [
I. 47] therefore the square on
AB is equal to the square on
RL; therefore
AB is equal to
RL.
But each of the straight lines
BC,
DE,
EF,
GH,
HK is equal to
AB, while each of the straight lines
RM,
RN is equal to
RL; therefore each of the straight lines
AB,
BC,
DE,
EF,
GH,
HK is equal to each of the straight lines
RL,
RM,
RN.
And, since the two sides
LR,
RM are equal to the two sides
AB,
BC, and the base
LM is by hypothesis equal to the base
AC, therefore the angle
LRM is equal to the angle
ABC. [
I. 8]
For the same reason the angle
MRN is also equal to the angle
DEF, and the angle
LRN to the angle
GHK.
Therefore, out of the three plane angles
LRM,
MRN,
LRN, which are equal to the three given angles
ABC,
DEF,
GHK, the solid angle at
R has been constructed, which is contained by the angles
LRM,
MRN,
LRN. Q. E. F.
LEMMA.
But how it is possible to take the square on
OR equal to that area by which the square on
AB is greater than the square on
LO, we can show as follows.
Let the straight lines
AB,
LO be set out, and let
AB be the greater; let the semicircle
ABC be described on
AB, and into the semicircle
ABC let
AC be fitted equal to the straight line
LO, not being greater than the diameter
AB; [
IV. 1] let
CB be joined
Since then the angle
ACB is an angle in the semicircle
ACB, therefore the angle
ACB is right. [
III. 31]
Therefore the square on
AB is equal to the squares on
AC,
CB. [
I. 47]
Hence the square on
AB is greater than the square on
AC by the square on
CB.
But
AC is equal to
LO.
Therefore the square on
AB is greater than the square on
LO by the square on
CB.
If then we cut off
OR equal to
BC, the square on
AB will be greater than the square on
LO by the square on
OR. Q. E. F.
PROPOSITION 24.
If a solid be contained by parallel planes,
the opposite planes in it are equal and parallelogrammic.
For let the solid
CDHG be contained by the parallel planes
AC,
GF,
AH,
DF,
BF,
AE; I say that the opposite planes in it are equal and parallelogrammic.
For, since the two parallel planes
BG,
CE are cut by the plane
AC, their common sections are parallel. [
XI. 16]
Therefore
AB is parallel to
DC.
Again, since the two parallel planes
BF,
AE are cut by the plane
AC, their common sections are parallel. [
XI. 16]
Therefore
BC is parallel to
AD.
But
AB was also proved parallel to
DC; therefore
AC is a parallelogram.
Similarly we can prove that each of the planes
DF,
FG,
GB,
BF,
AE is a parallelogram.
Let
AH,
DF be joined.
Then, since
AB is parallel to
DC, and
BH to
CF, the two straight lines
AB,
BH which meet one another are parallel to the two straight lines
DC,
CF which meet one another, not in the same plane; therefore they will contain equal angles; [
XI. 10] therefore the angle
ABH is equal to the angle
DCF.
And, since the two sides
AB,
BH are equal to the two sides
DC,
CF, [
I. 34] and the angle
ABH is equal to the angle
DCF, therefore the base
AH is equal to the base
DF, and the triangle
ABH is equal to the triangle
DCF. [
I. 4]
And the parallelogram
BG is double of the triangle
ABH, and the parallelogram
CE double of the triangle
DCF; [
I. 34] therefore the parallelogram
BG is equal to the parallelogram
CE.
Similarly we can prove that
AC is also equal to
GF, and
AE to
BF.
Therefore etc. Q. E. D.
PROPOSITION 25.
If a parallelepipedal solid be cut by a plane which is parallel to the opposite planes,
then,
as the base is to the base,
so will the solid be to the solid.
For let the parallelepipedal solid
ABCD be cut by the plane
FG which is parallel to the opposite planes
RA,
DH; I say that, as the base
AEFV is to the base
EHCF, so is the solid
ABFU to the solid
EGCD.
For let
AH be produced in each direction, let any number of straight lines whatever,
AK,
KL, be made equal to
AE, and any number whatever,
HM,
MN, equal to
EH; and let the parallelograms
LP,
KV,
HW,
MS and the solids
LQ,
KR,
DM,
MT be completed.
Then, since the straight lines
LK,
KA,
AE are equal to one another, the parallelograms
LP,
KV,
AF are also equal to one another,
KO,
KB,
AG are equal to one another, and further
LX,
KQ,
AR are equal to one another, for they are opposite. [
XI. 24]
For the same reason the parallelograms
EC,
HW,
MS are also equal to one another,
HG,
HI,
IN are equal to one another, and further
DH,
MY,
NT are equal to one another.
Therefore in the solids
LQ,
KR,
AU three planes are equal to three planes.
But the three planes are equal to the three opposite; therefore the three solids
LQ,
KR,
AU are equal to one another.
For the same reason the three solids
ED,
DM,
MT are also equal to one another.
Therefore, whatever multiple the base
LF is of the base
AF, the same multiple also is the solid
LU of the solid
AU.
For the same reason, whatever multiple the base
NF is of the base
FH, the same multiple also is the solid
NU of the solid
HU.
And, if the base
LF is equal to the base
NF, the solid
LU is also equal to the solid
NU; if the base
LF exceeds the base
NF, the solid
LU also exceeds the solid
NU; and, if one falls short, the other falls short.
Therefore, there being four magnitudes, the two bases
AF,
FH, and the two solids
AU,
UH, equimultiples have been taken of the base
AF and the solid
AU, namely the base
LF and the solid
LU, and equimultiples of the base
HF and the solid
HU, namely the base
NF and the solid
NU, and it has been proved that, if the base
LF exceeds the base
FN, the solid
LU also exceeds the solid
NU, if the bases are equal, the solids are equal, and if the base falls short, the solid falls short.
Therefore, as the base
AF is to the base
FH, so is the solid
AU to the solid
UH. [
V. Def. 5] Q. E. D.
PROPOSITION 26.
On a given straight line, and at a given point on it,
to construct a solid angle equal to a given solid angle.
Let
AB be the given straight line,
A the given point on it, and the angle at
D, contained by the angles
EDC,
EDF,
FDC, the given solid angle; thus it is required to construct on the straight line
AB, and at the point
A on it, a solid angle equal to the solid angle at
D.
For let a point
F be taken at random on
DF, let
FG be drawn from
F perpendicular to the plane through
ED,
DC, and let it meet the plane at
G, [
XI. 11] let
DG be joined, let there be constructed on the straight line
AB and at the point
A on it the angle
BAL equal to the angle
EDC, and the angle
BAK equal to the angle
EDG, [
I. 23] let
AK be made equal to
DG, let
KH be set up from the point
K at right angles to the plane through
BA,
AL, [
XI. 12] let
KH be made equal to
GF, and let
HA be joined; I say that the solid angle at
A, contained by the angles
BAL,
BAH,
HAL is equal to the solid angle at
D contained by the angles
EDC,
EDF,
FDC.
For let
AB,
DE be cut off equal to one another, and let
HB,
KB,
FE,
GE be joined.
Then, since
FG is at right angles to the plane of reference, it will also make right angles with all the straight lines which meet it and are in the plane of reference; [
XI. Def. 3] therefore each of the angles
FGD,
FGE is right.
For the same reason each of the angles
HKA,
HKB is also right.
And, since the two sides
KA,
AB are equal to the two sides
GD,
DE respectively, and they contain equal angles, therefore the base
KB is equal to the base
GE. [
I. 4]
But
KH is also equal to
GF, and they contain right angles; therefore
HB is also equal to
FE. [
I. 4]
Again, since the two sides
AK,
KH are equal to the two sides
DG,
GF, and they contain right angles, therefore the base
AH is equal to the base
FD. [
I. 4]
But
AB is also equal to
DE; therefore the two sides
HA,
AB are equal to the two sides
DF,
DE.
And the base
HB is equal to the base
FE; therefore the angle
BAH is equal to the angle
EDF. [
I. 8]
For the same reason the angle
HAL is also equal to the angle
FDC.
And the angle
BAL is also equal to the angle
EDC.
Therefore on the straight line
AB, and at the point
A on it, a solid angle has been constructed equal to the given solid angle at
D. Q. E. F.
PROPOSITION 27.
On a given straight line to describe a parallelepipedal solid similar and similarly situated to a given parallelepipedal solid.
Let
AB be the given straight line and
CD the given parallelepipedal solid; thus it is required to describe on the given straight line
AB a parallelepipedal solid similar and similarly situated to the given parallelepipedal solid
CD.
For on the straight line
AB and at the point
A on it let the solid angle, contained by the angles
BAH,
HAK,
KAB, be constructed equal to the solid angle at
C, so that the angle
BAH is equal to the angle
ECF, the angle
BAK equal to the angle
ECG, and the angle
KAH to the angle
GCF; and let it be contrived that, as
EC is to
CG, so is
BA to
AK, and, as
GC is to
CF, so is
KA to
AH. [
VI. 12]
Therefore also,
ex aequali, as
EC is to
CF, so is
BA to
AH. [
V. 22]
Let the parallelogram
HB and the solid
AL be completed.
Now since, as
EC is to
CG, so is
BA to
AK, and the sides about the equal angles
ECG,
BAK are thus proportional, therefore the parallelogram
GE is similar to the parallelogram
KB.
For the same reason the parallelogram
KH is also similar to the parallelogram
GF, and further
FE to
HB; therefore three parallelograms of the solid
CD are similar to three parallelograms of the solid
AL.
But the former three are both equal and similar to the three opposite parallelograms, and the latter three are both equal and similar to the three opposite parallelograms; therefore the whole solid
CD is similar to the whole solid
AL. [
XI. Def. 9]
Therefore on the given straight line
AB there has been described
AL similar and similarly situated to the given parallelepipedal solid
CD. Q. E. F.
PROPOSITION 28.
If a parallelepipedal solid be cut by a plane through the diagonals of the opposite planes,
the solid will be bisected by the plane.
For let the parallelepipedal solid
AB be cut by the plane
CDEF through the diagonals
CF,
DE of opposite planes; I say that the solid
AB will be bisected by the plane
CDEF.
For, since the triangle
CGF is equal to the triangle
CFB, [
I. 34] and
ADE to
DEH, while the parallelogram
CA is also equal to the parallelogram
EB, for they are opposite, and
GE to
CH, therefore the prism contained by the two triangles
CGF,
ADE and the three parallelograms
GE,
AC,
CE is also equal to the prism contained by the two triangles
CFB,
DEH and the three parallelograms
CH,
BE,
CE; for they are contained by planes equal both in multitude and in magnitude. [
XI. Def. 10]
Hence the whole solid
AB is bisected by the plane
CDEF. Q. E. D.
PROPOSITION 29.
Parallelepipedal solids which are on the same base and of the same height,
and in which the extremities of the sides which stand up are on the same straight lines,
are equal to one another.
Let
CM,
CN be parallelepipedal solids on the same base
AB and of the same height, and let the extremities of their sides which stand up, namely
AG,
AF,
LM,
LN,
CD,
CE,
BH,
BK, be on the same straight lines
FN,
DK; I say that the solid
CM is equal to the solid
CN.
For, since each of the figures
CH,
CK is a parallelogram,
CB is equal to each of the straight lines
DH,
EK, [
I. 34] hence
DH is also equal to
EK.
Let
EH be subtracted from each; therefore the remainder
DE is equal to the remainder
HK.
Hence the triangle
DCE is also equal to the triangle
HBK, [
I. 8, 4] and the parallelogram
DG to the parallelogram
HN. [
I. 36]
For the same reason the triangle
AFG is also equal to the triangle
MLN.
But the parallelogram
CF is equal to the parallelogram
BM, and
CG to
BN, for they are opposite; therefore the prism contained by the two triangles
AFG,
DCE and the three parallelograms
AD,
DG,
CG is equal to the prism contained by the two triangles
MLN,
HBK and the three parallelograms
BM,
HN,
BN.
Let there be added to each the solid of which the parallelogram
AB is the base and
GEHM its opposite; therefore the whole parallelepipedal solid
CM is equal to the whole parallelepipedal solid
CN.
Therefore etc. Q. E. D.
PROPOSITION 30.
Parallelepipedal solids which are on the same base and of the same height,
and in which the extremities of the sides which stand up are not on the same straight lines,
are equal to one another.
Let
CM,
CN be parallelepipedal solids on the same base
AB and of the same height, and let the extremities of their sides which stand up, namely
AF,
AG,
LM,
LN,
CD,
CE,
BH,
BK, not be on the same straight lines; I say that the solid
CM is equal to the solid
CN.
For let
NK,
DH be produced and meet one another at
R, and further let
FM,
GE be produced to
P,
Q; let
AO,
LP,
CQ,
BR be joined.
Then the solid
CM, of which the parallelogram
ACBL is the base, and
FDHM its opposite, is equal to the solid
CP, of which the parallelogram
ACBL is the base, and
OQRP its opposite; for they are on the same base
ACBL and of the same height, and the extremities of their sides which stand up, namely
AF,
AO,
LM,
LP,
CD,
CQ,
BH,
BR, are on the same straight lines
FP,
DR. [
XI. 29]
But the solid
CP, of which the parallelogram
ACBL is the base, and
OQRP its opposite, is equal to the solid
CN, of which the parallelogram
ACBL is the base and
GEKN its opposite; for they are again on the same base
ACBL and of the same height, and the extremities of their sides which stand up, namely
AG,
AO,
CE,
CQ,
LN,
LP,
BK,
BR, are on the same straight lines
GQ,
NR.
Hence the solid
CM is also equal to the solid
CN.
Therefore etc. Q. E. D.
PROPOSITION 31.
Parallelepipedal solids which are on equal bases and of the same height are equal to one another.
Let the parallelepipedal solids
AE,
CF, of the same height, be on equal bases
AB,
CD.
I say that the solid
AE is equal to the solid
CF.
First, let the sides which stand up,
HK,
BE,
AG,
LM,
PQ,
DF,
CO,
RS, be at right angles to the bases
AB,
CD; let the straight line
RT be produced in a straight line with
CR; on the straight line
RT, and at the point
R on it, let the angle
TRU be constructed equal to the angle
ALB, [
I. 23] let
RT be made equal to
AL, and
RU equal to
LB, and let the base
RW and the solid
XU be completed.
Now, since the two sides
TR,
RU are equal to the two sides
AL,
LB, and they contain equal angles, therefore the parallelogram
RW is equal and similar to the parallelogram
HL.
Since again
AL is equal to
RT, and
LM to
RS, and they contain right angles, therefore the parallelogram
RX is equal and similar to the parallelogram
AM.
For the same reason
LE is also equal and similar to
SU; therefore three parallelograms of the solid
AE are equal and similar to three parallelograms of the solid
XU.
But the former three are equal and similar to the three opposite, and the latter three to the three opposite; [
XI. 24] therefore the whole parallelepipedal solid
AE is equal to the whole parallelepipedal solid
XU. [
XI. Def. 10]
Let
DR,
WU be drawn through and meet one another at
Y, let
aTb be drawn through
T parallel to
DY, let
PD be produced to
a, and let the solids
YX,
RI be completed.
Then the solid
XY, of which the parallelogram
RX is the base and
Yc its opposite, is equal to the solid
XU of which the parallelogram
RX is the base and
UV its opposite, for they are on the same base
RX and of the same height, and the extremities of their sides which stand up, namely
RY,
RU,
Tb,
TW,
Se,
Sd,
Xc,
XV, are on the same straight lines
YW,
eV. [
XI. 29]
But the solid
XU is equal to
AE: therefore the solid
XY is also equal to the solid
AE.
And, since the parallelogram
RUWT is equal to the parallelogram
YT for they are on the same base
RT and in the same parallels
RT,
YW, [
I. 35] while
RUWT is equal to
CD, since it is also equal to
AB, therefore the parallelogram
YT is also equal to
CD.
But
DT is another parallelogram; therefore, as the base
CD is to
DT, so is
YT to
DT. [
V. 7]
And, since the parallelepipedal solid
CI has been cut by the plane
RF which is parallel to opposite planes, as the base
CD is to the base
DT, so is the solid
CF to the solid
RI. [
XI. 25]
For the same reason, since the parallelepipedal solid
YI has been cut by the plane
RX which is parallel to opposite planes, as the base
YT is to the base
TD, so is the solid
YX to the solid
RI. [
XI. 25]
But, as the base
CD is to
DT, so is
YT to
DT; therefore also, as the solid
CF is to the solid
RI, so is the solid
YX to
RI. [
V. 11]
Therefore each of the solids
CF,
YX has to
RI the same ratio; therefore the solid
CF is equal to the solid
YX. [
V. 9]
But
YX was proved equal to
AE; therefore
AE is also equal to
CF.
Next, let the sides standing up,
AG,
HK,
BE,
LM,
CN,
PQ,
DF,
RS, not be at right angles to the bases
AB,
CD; I say again that the solid
AE is equal to the solid
CF.
For from the points
K,
E,
G,
M,
Q,
F,
N,
S let
KO,
ET,
GU,
MV,
QW,
FX,
NY,
SI be drawn perpendicular to the plane of reference, and let them meet the plane at the points
O,
T,
U,
V,
W,
X,
Y,
I, and let
OT,
OU,
UV,
TV,
WX,
WY,
YI,
IX be joined.
Then the solid
KV is equal to the solid
QI, for they are on the equal bases
KM,
QS and of the same height, and their sides which stand up are at right angles to their bases. [First part of this Prop.]
But the solid
KV is equal to the solid
AE, and
QI to
CF; for they are on the same base and of the same height, while the extremities of their sides which stand up are not on the same straight lines. [
XI. 30]
Therefore the solid
AE is also equal to the solid
CF.
Therefore etc. Q. E. D.
PROPOSITION 32.
Parallelepipedal solids which are of the same height are to one another as their bases.
Let
AB,
CD be parallelepipedal solids of the same height; I say that the parallelepipedal solids
AB,
CD are to one another as their bases, that is, that, as the base
AE is to the base
CF, so is the solid
AB to the solid
CD.
For let
FH equal to
AE be applied to
FG, [
I. 45] and on
FH as base, and with the same height as that of
CD, let the parallelepipedal solid
GK be completed.
Then the solid
AB is equal to the solid
GK; for they are on equal bases
AE,
FH and of the same height. [
XI. 31]
And, since the parallelepipedal solid
CK is cut by the plane
DG which is parallel to opposite planes, therefore, as the base
CF is to the base
FH, so is the solid
CD to the solid
DH. [
XI. 25]
But the base
FH is equal to the base
AE, and the solid
GK to the solid
AB; therefore also, as the base
AE is to the base
CF, so is the solid
AB to the solid
CD.
Therefore etc. Q. E. D.
PROPOSITION 33.
Similar parallelepipedal solids are to one another in the triplicate ratio of their corresponding sides.
Let
AB,
CD be similar parallelepipedal solids, and let
AE be the side corresponding to
CF; I say that the solid
AB has to the solid
CD the ratio triplicate of that which
AE has to
CF.
For let
EK,
EL,
EM be produced in a straight line with
AE,
GE,
HE, let
EK be made equal to
CF,
EL equal to
FN, and further
EM equal to
FR, and let the parallelogram
KL and the solid
KP be completed.
Now, since the two sides
KE,
EL are equal to the two sides
CF,
FN, while the angle
KEL is also equal to the angle
CFN, inasmuch as the angle
AEG is also equal to the angle
CFN because of the similarity of the solids
AB,
CD, therefore the parallelogram
KL is equal <and similar> to the parallelogram
CN.
For the same reason the parallelogram
KM is also equal and similar to
CR, and further
EP to
DF; therefore three parallelograms of the solid
KP are equal and similar to three parallelograms of the solid
CD.
But the former three parallelograms are equal and similar to their opposites, and the latter three to their opposites; [
XI. 24] therefore the whole solid
KP is equal and similar to the whole solid
CD. [
XI. Def. 10]
Let the parallelogram
GK be completed, and on the parallelograms
GK,
KL as bases, and with the same height as that of
AB, let the solids
EO,
LQ be completed.
Then since; owing to the similarity of the solids
AB,
CD, as
AE is to
CF, so is
EG to
FN, and
EH to
FR, while
CF is equal to
EK,
FN to
EL, and
FR to
EM, therefore, as
AE is to
EK, so is
GE to
EL, and
HE to
EM.
But, as
AE is to
EK, so is
AG to the parallelogram
GK, as
GE is to
EL, so is
GK to
KL, and, as
HE is to
EM, so is
QE to
KM; [
VI. 1] therefore also, as the parallelogram
AG is to
GK, so is
GK to
KL, and
QE to
KM.
But, as
AG is to
GK, so is the solid
AB to the solid
EO, as
GK is to
KL, so is the solid
OE to the solid
QL, and, as
QE is to
KM, so is the solid
QL to the solid
KP; [
XI. 32] therefore also, as the solid
AB is to
EO, so is
EO to
QL, and
QL to
KP.
But, if four magnitudes be continuously proportional, the first has to the fourth the ratio triplicate of that which it has to the second; [
V. Def. 10] therefore the solid
AB has to
KP the ratio triplicate of that which
AB has to
EO.
But, as
AB is to
EO, so is the parallelogram
AG to
GK, and the straight line
AE to
EK [
VI. 1]; hence the solid
AB has also to
KP the ratio triplicate of that which
AE has to
EK.
But the solid
KP is equal to the solid
CD, and the straight line
EK to
CF; therefore the solid
AB has also to the solid
CD the ratio triplicate of that which the corresponding side of it,
AE, has to the corresponding side
CF.
Therefore etc. Q. E. D.
PORISM.
From this it is manifest that, if four straight lines be <continuously> proportional, as the first is to the fourth, so will a parallelepipedal solid on the first be to the similar and similarly described parallelepipedal solid on the second, inasmuch as the first has to the fourth the ratio triplicate of that which it has to the second.
PROPOSITION 34.
In equal parallelepipedal solids the bases are reciprocally proportional to the heights; and those parallelepipedal solids in which the bases are reciprocally proportional to the heights are equal.
Let
AB,
CD be equal parallelepipedal solids; I say that in the parallelepipedal solids
AB,
CD the bases are reciprocally proportional to the heights, that is, as the base
EH is to the base
NQ, so is the height of the solid
CD to the height of the solid
AB.
First, let the sides which stand up, namely
AG,
EF,
LB,
HK,
CM,
NO,
PD,
QR, be at right angles to their bases; I say that, as the base
EH is to the base
NQ, so is
CM to
AG.
If now the base
EH is equal to the base
NQ, while the solid
AB is also equal to the solid
CD,
CM will also be equal to
AG.
For parallelepipedal solids of the same height are to one another as the bases; [
XI. 32] and, as the base
EH is to
NQ, so will
CM be to
AG, and it is manifest that in the parallelepipedal solids
AB,
CD the bases are reciprocally proportional to the heights.
Next, let the base
EH not be equal to the base
NQ, but let
EH be greater.
Now the solid
AB is equal to the solid
CD; therefore
CM is also greater than
AG.
Let then
CT be made equal to
AG, and let the parallelepipedal solid
VC be completed on
NQ as base and with
CT as height.
Now, since the solid
AB is equal to the solid
CD, and
CV is outside them, while equals have to the same the same ratio, [
V. 7] therefore, as the solid
AB is to the solid
CV, so is the solid
CD to the solid
CV.
But, as the solid
AB is to the solid
CV, so is the base
EH to the base
NQ, for the solids
AB,
CV are of equal height; [
XI. 32] and, as the solid
CD is to the solid
CV, so is the base
MQ to the base
TQ [
XI. 25] and
CM to
CT [
VI. 1]; therefore also, as the base
EH is to the base
NQ, so is
MC to
CT.
But
CT is equal to
AG; therefore also, as the base
EH is to the base
NQ, so is
MC to
AG.
Therefore in the parallelepipedal solids
AB,
CD the bases are reciprocally proportional to the heights.
Again, in the parallelepipedal solids
AB,
CD let the bases be reciprocally proportional to the heights, that is, as the base
EH is to the base
NQ, so let the height of the solid
CD be to the height of the solid
AB; I say that the solid
AB is equal to the solid
CD.
Let the sides which stand up be again at right angles to the bases.
Now, if the base
EH is equal to the base
NQ, and, as the base
EH is to the base
NQ, so is the height of the solid
CD to the height of the solid
AB, therefore the height of the solid
CD is also equal to the height of the solid
AB.
But parallelepipedal solids on equal bases and of the same height are equal to one another; [
XI. 31] therefore the solid
AB is equal to the solid
CD.
Next, let the base
EH not be equal to the base
NQ, but let
EH be greater; therefore the height of the solid
CD is also greater than the height of the solid
AB, that is,
CM is greater than
AG.
Let
CT be again made equal to
AG, and let the solid
CV be similarly completed.
Since, as the base
EH is to the base
NQ, so is
MC to
AG, while
AG is equal to
CT, therefore, as the base
EH is to the base
NQ, so is
CM to
CT.
But, as the base
EH is to the base
NQ, so is the solid
AB to the solid
CV, for the solids
AB,
CV are of equal height; [
XI. 32] and, as
CM is to
CT, so is the base
MQ to the base
QT [
VI. 1] and the solid
CD to the solid
CV. [
XI. 25]
Therefore also, as the solid
AB is to the solid
CV, so is the solid
CD to the solid
CV; therefore each of the solids
AB,
CD has to
CV the same ratio.
Therefore the solid
AB is equal to the solid
CD. [
V. 9]
Now let the sides which stand up,
FE,
BL,
GA,
HK,
ON,
DP,
MC,
RQ, not be at right angles to their bases; let perpendiculars be drawn from the points
F,
G,
B,
K,
O,
M,
D,
R to the planes through
EH,
NQ, and let them meet the planes at
S,
T,
U,
V,
W,
X,
Y,
a, and let the solids
FV,
Oa be completed; I say that, in this case too, if the solids
AB,
CD are equal, the bases are reciprocally proportional to the heights, that is, as the base
EH is to the base
NQ, so is the height of the solid
CD to the height of the solid
AB.
Since the solid
AB is equal to the solid
CD, while
AB is equal to
BT, for they are on the same base
FK and of the same height; [
XI. 29, 30] and the solid
CD is equal to
DX, for they are again on the same base
RO and of the same height; [
id.] therefore the solid
BT is also equal to the solid
DX.
Therefore, as the base
FK is to the base
OR, so is the height of the solid
DX to the height of the solid
BT. [Part 1.]
But the base
FK is equal to the base
EH, and the base
OR to the base
NQ; therefore, as the base
EH is to the base
NQ, so is the height of the solid
DX to the height of the solid
BT.
But the solids
DX,
BT and the solids
DC,
BA have the same heights respectively; therefore, as the base
EH is to the base
NQ, so is the height of the solid
DC to the height of the solid
AB.
Therefore in the parallelepipedal solids
AB,
CD the bases are reciprocally proportional to the heights.
Again, in the parallelepipedal solids
AB,
CD let the bases be reciprocally proportional to the heights, that is, as the base
EH is to the base
NQ, so let the height of the solid
CD be to the height of the solid
AB; I say that the solid
AB is equal to the solid
CD.
For, with the same construction, since, as the base
EH is to the base
NQ, so is the height of the solid
CD to the height of the solid
AB, while the base
EH is equal to the base
FK, and
NQ to
OR, therefore, as the base
FK is to the base
OR, so is the height of the solid
CD to the height of the solid
AB.
But the solids
AB,
CD and
BT,
DX have the same heights respectively; therefore, as the base
FK is to the base
OR, so is the height of the solid
DX to the height of the solid
BT.
Therefore in the parallelepipedal solids
BT,
DX the bases are reciprocally proportional to the heights; therefore the solid
BT is equal to the solid
DX. [Part 1.]
But
BT is equal to
BA, for they are on the same base
FK and of the same height; [
XI. 29, 30] and the solid
DX is equal to the solid
DC. [
id.]
Therefore the solid
AB is also equal to the solid
CD. Q. E. D.
PROPOSITION 35.
If there be two equal plane angles,
and on their vertices there be set up elevated straight lines containing equal angles with the original straight lines respectively,
if on the elevated straight lines points be taken at random and perpendiculars be drawn from them to the planes in which the original angles are,
and if from the points so arising in the planes straight lines be joined to the vertices of the original angles,
they will contain,
with the elevated straight lines,
equal angles.
Let the angles
BAC,
EDF be two equal rectilineal angles, and from the points
A,
D let the elevated straight lines
AG,
DM be set up containing, with the original straight lines, equal angles respectively, namely, the angle
MDE to the angle
GAB and the angle
MDF to the angle
GAC, let points
G,
M be taken at random on
AG,
DM, let
GL,
MN be drawn from the points
G,
M perpendicular to the planes through
BA,
AC and
ED,
DF, and let them meet the planes at
L,
N, and let
LA,
ND be joined; I say that the angle
GAL is equal to the angle
MDN.
Let
AH be made equal to
DM, and let
HK be drawn through the point
H parallel to
GL.
But
GL is perpendicular to the plane through
BA,
AC; therefore
HK is also perpendicular to the plane through.
BA,
AC. [
XI. 8]
From the points
K,
N let
KC,
NF,
KB,
NE be drawn perpendicular to the straight lines
AC,
DF,
AB,
DE, and let
HC,
CB,
MF,
FE be joined.
Since the square on
HA is equal to the squares on
HK,
KA, and the squares on
KC,
CA are equal to the square on
KA, [
I. 47] therefore the square on
HA is also equal to the squares on
HK,
KC,
CA.
But the square on
HC is equal to the squares on
HK,
KC; [
I. 47] therefore the square on
HA is equal to the squares on
HC,
CA.
Therefore the angle
HCA is right. [
I. 48]
For the same reason the angle
DFM is also right.
Therefore the angle
ACH is equal to the angle
DFM.
But the angle
HAC is also equal to the angle
MDF.
Therefore
MDF,
HAC are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely, that subtending one of the equal angles, that is,
HA equal to
MD; therefore they will also have the remaining sides equal to the remaining sides respectively. [
I. 26]
Therefore
AC is equal to
DF.
Similarly we can prove that
AB is also equal to
DE.
Since then
AC is equal to
DF, and
AB to
DE, the two sides
CA,
AB are equal to the two sides
FD,
DE.
But the angle
CAB is also equal to the angle
FDE; therefore the base
BC is equal to the base
EF, the triangle to the triangle, and the remaining angles to the remaining angles; [
I. 4] therefore the angle
ACB is equal to the angle
DFE.
But the right angle
ACK is also equal to the right angle
DFN; therefore the remaining angle
BCK is also equal to the remaining angle
EFN.
For the same reason the angle
CBK is also equal to the angle
FEN.
Therefore
BCK,
EFN are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely, that adjacent to the equal angles, that is,
BC equal to
EF; therefore they will also have the remaining sides equal to the remaining sides. [
I. 26]
Therefore
CK is equal to
FN.
But
AC is also equal to
DF; therefore the two sides
AC,
CK are equal to the two sides
DF,
FN; and they contain right angles.
Therefore the base
AK is equal to the base
DN. [
I. 4]
And, since
AH is equal to
DM, the square on
AH is also equal to the square on
DM.
But the squares on
AK,
KH are equal to the square on
AH, for the angle
AKH is right; [
I. 47] and the squares on
DN,
NM are equal to the square on
DM, for the angle
DNM is right; [
I. 47] therefore the squares on
AK,
KH are equal to the squares on
DN,
NM; and of these the square on
AK is equal to the square on
DN; therefore the remaining square on
KH is equal to the square on
NM; therefore
HK is equal to
MN.
And, since the two sides
HA,
AK are equal to the two sides
MD,
DN respectively, and the base
HK was proved equal to the base
MN, therefore the angle
HAK is equal to the angle
MDN. [
I. 8]
Therefore etc.
PORISM.
From this it is manifest that, if there be two equal plane angles, and if there be set up on them elevated straight lines which are equal and contain equal angles with the original straight lines respectively, the perpendiculars drawn from their extremities to the planes in which are the original angles are equal to one another. Q. E. D.
PROPOSITION 36.
If three straight lines be proportional,
the parallelepipedal solid formed out of the three is equal to the parallelepipedal solid on the mean which is equilateral,
but equiangular with the aforesaid solid.
Let
A,
B,
C be three straight lines in proportion, so that, as
A is to
B, so is
B to
C; I say that the solid formed out of
A,
B,
C is equal to the solid on
B which is equilateral, but equiangular with the aforesaid solid.
Let there be set out the solid angle at
E contained by the angles
DEG,
GEF,
FED, let each of the straight lines
DE,
GE,
EF be made equal to
B, and let the parallelepipedal solid
EK be completed, let
LM be made equal to
A, and on the straight line
LM, and at the point
L on it, let there be constructed a solid angle equal to the solid angle at
E, namely that contained by
NLO,
OLM,
MLN; let
LO be made equal to
B, and
LN equal to
C.
Now, since, as
A is to
B, so is
B to
C, while
A is equal to
LM,
B to each of the straight lines
LO,
ED, and
C to
LN, therefore, as
LM is to
EF, so is
DE to
LN.
Thus the sides about the equal angles
NLM,
DEF are reciprocally proportional; therefore the parallelogram
MN is equal to the parallelogram
DF. [
VI. 14]
And, since the angles
DEF,
NLM are two plane rectilineal angles, and on them the elevated straight lines
LO,
EG are set up which are equal to one another and contain equal angles with the original straight lines respectively, therefore the perpendiculars drawn from the points
G,
O to the planes through
NL,
LM and
DE,
EF are equal to one another; [
XI. 35, Por.] hence the solids
LH,
EK are of the same height.
But parallelepipedal solids on equal bases and of the same height are equal to one another; [
XI. 31] therefore the solid
HL is equal to the solid
EK.
And
LH is the solid formed out of
A,
B,
C, and
EK the solid on
B; therefore the parallelepipedal solid formed out of
A,
B,
C is equal to the solid on
B which is equilateral, but equiangular with the aforesaid solid. Q. E. D.
PROPOSITION 37.
If four straight lines be proportional,
the parallelepipedal solids on them which are similar and similarly described will also be proportional; and,
if the parallelepipedal solids on them which are similar and similarly described be proportional,
the straight lines will themselves also be proportional.
Let
AB,
CD,
EF,
GH be four straight lines in proportion, so that, as
AB is to
CD, so is
EF to
GH; and let there be described on
AB,
CD,
EF,
GH the similar and similarly situated parallelepipedal solids
KA,
LC,
ME,
NG; I say that, as
KA is to
LC, so is
ME to
NG.
For, since the parallelepipedal solid
KA is similar to
LC, therefore
KA has to
LC the ratio triplicate of that which
AB has to
CD. [
XI. 33]
For the same reason
ME also has to
NG the ratio triplicate of that which
EF has to
GH. [
id.]
And, as
AB is to
CD, so is
EF to
GH.
Therefore also, as
AK is to
LC, so is
ME to
NG.
Next, as the solid
AK is to the solid
LC, so let the solid
ME be to the solid
NG; I say that, as the straight line
AB is to
CD, so is
EF to
GH.
For since, again,
KA has to
LC the ratio triplicate of that which
AB has to
CD, [
XI. 33] and
ME also has to
NG the ratio triplicate of that which
EF has to
GH, [
id.] and, as
KA is to
LC, so is
ME to
NG, therefore also, as
AB is to
CD, so is
EF to
GH.
Therefore etc. Q. E. D.
PROPOSITION 38.
If the sides of the opposite planes of a cube be bisected,
and planes be carried through the points of section,
the common section of the planes and the diameter of the cube bisect one another.
For let the sides of the opposite planes
CF,
AH of the cube
AF be bisected at the points
K,
L,
M,
N,
O,
Q,
P,
R, and through the points of section let the planes
KN,
OR be carried; let
US be the common section of the planes, and
DG the diameter of the cube
AF.
I say that
UT is equal to
TS, and
DT to
TG.
For let
DU,
UE,
BS,
SG be joined.
Then, since
DO is parallel to
PE, the alternate angles
DOU,
UPE are equal to one another. [
I. 29]
And, since
DO is equal to
PE, and
OU to
UP, and they contain equal angles, therefore the base
DU is equal to the base
UE, the triangle
DOU is equal to the triangle
PUE, and the remaining angles are equal to the remaining angles; [
I. 4] therefore the angle
OUD is equal to the angle
PUE.
For this reason
DUE is a straight line. [
I. 14]
For the same reason,
BSG is also a straight line, and
BS is equal to
SG.
Now, since
CA is equal and parallel to
DB, while
CA is also equal and parallel to
EG, therefore
DB is also equal and parallel to
EG. [
XI. 9]
And the straight lines
DE,
BG join their extremities; therefore
DE is parallel to
BG. [
I. 33]
Therefore the angle
EDT is equal to the angle
BGT, for they are alternate; [
I. 29] and the angle
DTU is equal to the angle
GTS. [
I. 15]
Therefore
DTU,
GTS are two triangles which have two angles equal to two angles, and one side equal to one side, namely that subtending one of the equal angles, that is,
DU equal to
GS, for they are the halves of
DE,
BG; therefore they will also have the remaining sides equal to the remaining sides. [
I. 26]
Therefore
DT is equal to
TG, and
UT to
TS.
Therefore etc. Q. E. D.
PROPOSITION 39.
If there be two prisms of equal height,
and one have a parallelogram as base and the other a triangle,
and if the parallelogram be double of the triangle,
the prisms will be equal.
Let
ABCDEF,
GHKLMN be two prisms of equal height, let one have the parallelogram
AF as base, and the other the triangle
GHK, and let the parallelogram
AF be double of the triangle
GHK; I say that the prism
ABCDEF is equal to the prism
GHKLMN.
For let the solids
AO,
GP be completed.
Since the parallelogram
AF is double of the triangle
GHK, while the parallelogram
HK is also double of the triangle
GHK, [
I. 34] therefore the parallelogram
AF is equal to the parallelogram
HK.
But parallelepipedal solids which are on equal bases and of the same height are equal to one another; [
XI. 31] therefore the solid
AO is equal to the solid
GP.
And the prism
ABCDEF is half of the solid
AO, and the prism
GHKLMN is half of the solid
GP; [
XI. 28] therefore the prism
ABCDEF is equal to the prism
GHKLMN.
Therefore etc. Q. E. D.