Construction of Parallelepiped Similar to Given Parallelepiped

Proof

 * Euclid-XI-27.png

Let $AB$ be the given straight line.

Let $CD$ be the given parallelepiped.

Using :
 * 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 :
 * Let it be contrived that:
 * $EC : CG = BA : AK$

and:
 * $GC : CF = KA : AH$

Therefore from :
 * $EC : CF = BA : AH$

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 $AL$.