# Parallelepipeds on Equal Bases and Same Height are Equal in Volume

## Theorem

In the words of Euclid:

*Parallelepidedal solids which are on equal bases and of the same height are equal to one another.*

(*The Elements*: Book $\text{XI}$: Proposition $31$)

## 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 Proposition $23$ of Book $\text{I} $: Construction of Equal Angle:

- 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 $XU$.

- the former three are equal and similar to the three opposite.

Therefore from Book $\text{XI}$ Definition $10$: Similar Equal Solid Figures:

- 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 Proposition $29$ of Book $\text{XI} $: Parallelepipeds on Same Base and Same Height whose Extremities are on Same Lines are Equal in Volume:

- 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 Proposition $35$ of Book $\text{I} $: Parallelograms with Same Base and Same Height have Equal Area:

- the parallelogram $RUWT$ equals the parallelogram $YT$.

But $RUWT = AB$.

Therefore $RUWT = CD$

Therefore $YT = CD$.

But $DT$ is another parallelogram.

Therefore from Proposition $7$ of Book $\text{V} $: Ratios of Equal Magnitudes:

- 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 Proposition $25$ of Book $\text{XI} $: Parallelepiped cut by Plane Parallel to Opposite Planes:

- 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 Proposition $25$ of Book $\text{XI} $: Parallelepiped cut by Plane Parallel to Opposite Planes:

- 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 $AE = CF$.

$\Box$

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 $QI$.

- the parallelepiped $KV$ equals the parallelepiped $AE$

and:

- the parallelepiped $QI$ equals the parallelepiped $CF$.

Therefore the parallelepiped $AE$ is equal to the parallelepiped $CF$.

$\blacksquare$

## Historical Note

This proof is Proposition $31$ of Book $\text{XI}$ of Euclid's *The Elements*.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 3*(2nd ed.) ... (previous) ... (next): Book $\text{XI}$. Propositions