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

Proof

 * Euclid-XI-34.png

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 :
 * $AB = CD$


 * Euclid-XI-34-2.png

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:
 * $AB = CD$