Parallelepipeds on Equal Bases and Same Height are Equal in Volume

== Theorem ==
== Proof ==
:
Let $AE$ and $CF$ be parallelepipeds of the same height on equal bases $AB$ and $CD$.
It is to be demonstrated that the parallelepiped $AE$ is equal to the parallelepiped $CF$.
Let the sides $HK, BE, AG, LM, PQ, DF, CO, RS$ be perpendicular to the bases $AB$ and $CD$.
Let the straight line $RT$ be produced in a straight line with $CR$.
From :
:let $\angle TRU$ be constructed on $RT$ at $R$ equal to $\angle ALB$.
Let $RT = AL$.
Let $RU = LB$.
Let the base $RW$ and the parallelepiped $XU$ be completed.
We have that the two sides $TR$ and $RU$ are equal to the two sides $AL$ and $LB$.
$TR, RU$ and $AL, LB$ contain equal angles.
Therefore the parallelogram $RW$ is equal and similar to the parallelogram $HL$.
We have that $AL = RT$ and $LM = RS$.
$AL, LM$ and $RT, RS$ contain right angles.
Therefore the parallelogram $RX$ is equal and similar to the parallelogram $AM$.
For the same reason, the parallelogram $LE$ is equal and similar to the parallelogram $SU$.
Therefore three parallelograms of the parallelepiped $AE$ are equal and similar to three parallelograms of the parallelepiped $XU$.
But from :
:the former three are equal and similar to the three opposite.
Therefore from :
:the whole parallelepiped $AE$ is equal to the whole parallelepiped $XU$.
Let $DR$ and $WU$ be drawn through and meet one another at $Y$.
Let $A'TB'$ be drawn through $T$ parallel to $DY$.
Let $PD$ be produced to $A'$.
Let the parallelepipeds $YX$ and $RI$ be completed.
The parallelepiped $YX$ has base $RX$ whose opposite is $YC'$.
Then $YX$ equals the parallelepiped $XU$ which has a base $RX$ whose opposite is $UV$.
It follows from :
:the parallelepiped $XU$ equals the parallelepiped $XY$.
But $XU = AE$.
Therefore the parallelepiped $XY$ equals the parallelepiped $AE$.
We have that the parallelogram $RUWT$ is on the same base as the parallelogram $YT$.
Also, $RUWT$ is between the same parallels as $YT$.
Therefore from :
:the parallelogram $RUWT$ equals the parallelogram $YT$.
But $RUWT = AB$.
Therefore $RUWT = CD$
Therefore $YT = CD$.
But $DT$ is another parallelogram.
Therefore from :
:as the base $CD$ is to $DT$, so is $YT$ to $DT$.
We have that the parallelepiped $CI$ has been cut by the plane $RF$ which is parallel to opposite planes.
So from :
:as the base $CD$ is to $DT$, so is the parallelepiped $CF$ to the parallelepiped $RI$.
For the same reason, as the parallelepiped $YI$ has been cut by the plane $RX$ which is parallel to opposite planes,
from :
:as the base $YT$ is to $TD$, so is the parallelepiped $YX$ to the parallelepiped $RI$.
But as the base $CD$ is to $DT$, so is $YT$ is to $TD$.
Therefore, also, as the parallelepiped $CF$ is to the parallelepiped $RI$, so is the parallelepiped $YX$ to $RI$.
Therefore each of the parallelepipeds $CF$ and $YX$ has to $RI$ the same ratio.
Therefore the parallelepiped $CF$ equals the parallelepiped $YX$.
But parallelepiped $YX$ was proved equal to $AE$.
Therefore $AE = CF$.
:
Next, suppose the sides $AG, HK, BE, LM, CN, PQ, DF, RS$ not be perpendicular to the bases $AB$ and $CD$.
It is to be demonstrated that the parallelepiped $AE$ is equal to the parallelepiped $CF$.
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.
Let them meet the plane at the points $O, T, U, V, W, X, Y, I$.
Let $OT, OU, UV, TV, X, WY, YI, IX$ be joined.
Then from the first part of this proposition, the parallelepiped $KV$ equals the parallelepiped $QI$.
But from :
:the parallelepiped $KV$ equals the parallelepiped $AE$
and:
:the parallelepiped $QI$ equals the parallelepiped $CF$.
Therefore the parallelepiped $AE$ is equal to the parallelepiped $CF$.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Parallelepipeds