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.