Parallelepiped cut by Plane Parallel to Opposite Planes

Theorem

In the words of Euclid:

If a parallelepipedal solid be cut by a plane which is parallel to the opposite planes, then, as the base is to the base, so will the solid be to the solid.

(The Elements: Book $\text{XI}$: Proposition $25$)

Proof

Let the parallelepiped $ABCD$ be cut by the plane $FG$ which is parallel to the opposite planes $RA$ and $DH$.

It is to be demonstrated that the ratio of the base $AEFV$ to the base $EHCF$ equals the ratio of the parallelepiped $ABFU$ to the parallelepiped $EGCD$.

Let $AH$ be produced in each direction.

Let any number of straight lines $AK, KL$ equal to $AE$ be made.

Let any number of straight lines $HM, MN$ equal to $EH$ be made.

Let the parallelograms $LP, KV, HW, MS$ be completed.

Let the parallelepipeds $LQ, KR, DM, MT$ be completed.

We have that the straight lines $LK, KA, AE$ are equal.

Therefore the parallelograms $LP, KV, AF$ are equal.

From Proposition $24$ of Book $\text{XI} $: Opposite Planes of Solid contained by Parallel Planes are Equal Parallelograms:

the parallelograms $LX, KQ, AR$ are equal.

For the same reason, the parallelograms $EC, HW, MS$ are equal.

Therefore the parallelepipeds $HG, HI, NT$ are equal.

Further, the parallelograms $DH, MY, MT$ are equal.

Therefore, in the parallelepipeds $LQ, KR, AU$, three planes equal three planes.

But the three planes equal the three planes opposite.

Therefore the three parallelepipeds $LQ, KR, AU$ are equal to one another.

For the same reason, the three parallelepipeds $ED, DM, MT$ are equal to one another.

Therefore, whatever multiple the base $LF$ is of the base $AF$, the same multiple also is the parallelepiped $LU$ of the parallelepiped $AU$.

For the same reason, whatever multiple the base $NF$ is of the base $FH$, the same multiple also is the parallelepiped $NU$ of the parallelepiped $HU$.

And if the base $LF$ equals the base $NF$, the parallelepiped $LU$ equals the parallelepiped $NU$.

If the base $LF$ exceeds the base $NF$, the parallelepiped $LU$ exceeds the parallelepiped $NU$.

If the base $LF$ falls short of the base $NF$, the parallelepiped $LU$ falls short of the parallelepiped $NU$.

We have four magnitudes:

the two bases $AF$ and $FH$

and:

the two parallelepipeds $AU$ and $HU$.

Equimultiples have been taken of the base $AF$ and the parallelepiped $AU$: the base $LF$ and the parallelepiped $LU$.

and equimultiples have been taken of the base $HF$ and the parallelepiped $HU$: the base $NF$ and the parallelepiped $NU$.

It has been proved that:

if the base $LF$ exceeds the base $FN$, the parallelepiped $LU$ exceeds the parallelepiped $NU$

if the base $LF$ equals the base $NF$, the parallelepiped $LU$ equals the parallelepiped $NU$

if the base $LF$ falls short of the base $NF$, the parallelepiped $LU$ falls short of the parallelepiped $NU$.

Therefore, as the base $AF$ is to the base $FH$, so the parallelepiped $AU$ is to the parallelepiped $UH$.

$\blacksquare$

Historical Note

This proof is Proposition $25$ 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