Parallelepipeds are of Equal Volume iff Bases are in Reciprocal Proportion to Heights

== Theorem ==
== Proof ==
:
Let $AB$ and $CD$ be equal parallelepipeds.
It is to be demonstrated that the bases of $AB$ and $CD$ are inversely proportional to their heights.
That is:
:$EH : NQ = CM : AG$
First, let $AG, EF, LB, HK, CM, NO, PD, QR$ be perpendicular to their bases.
It is to be shown that:
:$EH : NQ = CM : AG$
Suppose $EH = NQ$.
From :
:$CM = AG$
Thus it follows trivially that:
:$EH : NQ = CM : AG$
Next, suppose $EH \ne NQ$.
Suppose WLOG that $EH > NQ$.
We have that $AB = CD$.
Therefore $CM > AG$.
Let $CT = AG$.
Let the parallelepiped $CV$ be completed on the base $NQ$ and with height $CT$.
We have that $AB = CD$.
So from :
:$AB : CV = CD : CV$
But from :
:$AB : CV = EH : NQ$
From :
:$CD : CV = MQ : TQ$
and from :
:$CD : CV = CM : CT$
Therefore also:
:$EH : NQ = MC : CT$
But:
:$CT = AG$
Therefore also:
:$EH : NQ = MC : AG$
Therefore in the parallelepipeds $AB$ and $CD$, the bases are inversely proportional to their heights.
Now let the bases of the parallelepipeds $AB$ and $CD$ be inversely proportional to their heights.
That is:
:$EH : NQ = MC : AG$
It is to be shown that $AB = CD$.
Let the sides $AG, EF, LB, HK, CM, NO, PD, QR$ be perpendicular to their bases.
Let $EH = NQ$.
Then it follows from $EH : NQ = MC : AG$ that $MC = AG$.
From :
:$AB = CD$
Next suppose that $EH \ne NQ$.
Suppose WLOG that $EH > NQ$.
Therefore the height of $CD$ is greater than the height of $AB$.
That is:
:$CM > AG$
Let $CT = AG$.
Let the parallelepiped $CV$ be completed on the base $NQ$ and with height $CT$.
We have that:
:$EH : NQ = MC : AG$
while:
:$AG = CT$
Therefore:
:$EH : NQ = CM : CT$
We have that $AB$ and $CV$ are of equal height.
So from :
:$EH : NQ = AB : CV$
From :
:$CM : CT = MQ : QT$
and from :
:$CM : CT = CD : CV$
Therefore also:
:$AB : CV = CD : CV$
Therefore from :
:$AB = CD$
:
Now let $AG, EF, LB, HK, CM, NO, PD, QR$ not be perpendicular to their bases.
Let perpendiculars be drawn from $F, G, B, K, O, M, D, R$ to the planes through $EH$ and $NQ$, meeting them at $S, T, U, V, W, X, Y, A'$.
Let the parallelepipeds $FV$ and $OA'$ be completed.
We have that:
:$AB = CD$
We have that $AB$ and $BT$ are on the same base $FK$ and have the same height.
From:
:
and:
::
we have that:
:$AB = BT$
For the same reason:
:$CD = DX$
Therefore:
:$BT = DX$
Therefore from the first part of this result:
:$FK : OR$ is the same as the ratio of the heights of $DX$ and $BT$.
But:
:$FK = EH$
and:
:$OR = NQ$
Therefore:
:$EH : NQ$ is the same as the ratio of the heights of $DX$ and $BT$.
But $DX$ and $BT$ have the same heights respectively as $DC$ and $BA$.
Therefore:
:$EH : NQ$ is the same as the ratio of the heights of $DC$ and $BA$.
Therefore in the parallelepipeds $AB$ and $CD$, the bases are inversely proportional to their heights.
Now let the bases of the parallelepipeds $AB$ and $CD$ be inversely proportional to their heights.
Let perpendiculars be drawn from $F, G, B, K, O, M, D, R$ to the planes through $EH$ and $NQ$, meeting them at $S, T, U, V, W, X, Y, A'$.
Let the parallelepipeds $FV$ and $OA'$ be completed.
Then as $DY$ and $BU$ are the heights of $AB$ and $CD$:
:$EH : NQ = DY : BU$
Thus:
:$EH : NQ$ equals the ratio of the heights of $DC$ and $BA$.
We have:
:$EH = FK$
and:
:$NQ = OR$
Therefore:
:$FK : OR$ equals the ratio of the heights of $DC$ and $BA$.
But $AB$ and $CD$ have the same heights as $BT$ and $DX$ respectively.
Therefore:
:$FK : OR$ equals the ratio of the heights of $DX$ and $BT$.
Therefore the bases of the parallelepipeds $BT$ and $DX$ are inversely proportional to their heights.
Therefore by the first part of this result:
:$BT = DX$
But from:
:
and:
::
we have that:
:$AB = BT$
and:
:$DX = DC$
Therefore:
:$AB = CD$
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Parallelepipeds