Construction of Parallelepiped Similar to Given Parallelepiped

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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

Euclid-XI-27.png

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