Construction of Parallelepiped Similar to Given Parallelepiped
Theorem
In the words of Euclid:
- On a given straight line to describe a parallelepidedal solid similar and similarly situated to a given parallelepidedal solid.
(The Elements: Book $\text{XI}$: Proposition $27$)
Proof
Let $AB$ be the given straight line.
Let $CD$ be the given parallelepiped.
Using Proposition $26$ of Book $\text{XI} $: Construction of Solid Angle equal to Given Solid Angle:
- On the straight line $AB$ let a solid angle contained by the plane angles $\angle BAH, \angle HAK, \angle KAB$ be constructed equal to the solid angle at $C$ contained by the plane angles $\angle ECF, \angle ECG, \angle GCF$ such that:
- $\angle BAH = \angle ECF$
- $\angle KAH = \angle GCF$
- $\angle BAK = \angle ECG$
Using Proposition $12$ of Book $\text{VI} $: Construction of Fourth Proportional Straight Line:
- Let it be contrived that:
- $EC : CG = BA : AK$
and:
- $GC : CF = KA : AH$
Therefore from Proposition $12$ of Book $\text{V} $: Equality of Ratios Ex Aequali:
- $EC : CF = BA : AH$
Let the parallelogram $HB$ and the parallelepiped $AL$ be completed.
We have that:
- $EC : CG = BA : AK$
and the sides about the equal angles $\angle BAK$ and $\angle ECG$ are proportional.
Therefore the parallelogram $GE$ is similar to the parallelogram $KB$.
For the same reason:
- the parallelogram $KH$ is similar to the parallelogram $GF$.
- the parallelogram $FE$ is similar to the parallelogram $HB$.
Therefore three parallelograms of the parallelepiped $CD$ are similar to the three parallelograms of the parallelepiped $AL$.
But the former three are equal and similar to the three opposite parallelograms.
Therefore the whole parallelepiped $CD$ is similar to the whole parallelepiped $AL$.
$\blacksquare$
Historical Note
This proof is Proposition $27$ 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