Proof: By Euclid

Let the given straight line be $AB$, and the given parallelepipedal solid $CD$.
So, it is necessary to describe a parallelepipedal solid similar, and similarly laid out, to the given parallelepipedal solid $CD$ on the given straight line $AB$.

fig27e

For, let a (solid angle) contained by the (rectilinear angles) $BAH$, $HAK$, and $KAB$ have been constructed, equal to solid angle at $C$, on the straight line $AB$ at the point $A$ on it [Prop. 11.26], such that angle $BAH$ is equal to $ECF$, and $BAK$ to $ECG$, and $KAH$ to $GCF$.
And let it have been contrived that as $EC$ (is) to $CG$, so $BA$ (is) to $AK$, and as $GC$ (is) to $CF$, so $KA$ (is) to $AH$ [Prop. 6.12].
And thus, via equality, as $EC$ is to $CF$, so $BA$ (is) to $AH$ [Prop. 5.22].
And let the parallelogram $HB$ have been completed, and the solid $AL$.
And since as $EC$ is to $CG$, so $BA$ (is) to $AK$, and the sides about the equal angles $ECG$ and $BAK$ are (thus) proportional, the parallelogram $GE$ is thus similar to the parallelogram $KB$.
So, for the same (reasons), the parallelogram $KH$ is also similar to the parallelogram $GF$, and, further, $FE$ (is similar) to $HB$.
Thus, three of the parallelograms of solid $CD$ are similar to three of the parallelograms of solid $AL$.
But, the (former) three are equal and similar to the three opposite, and the (latter) three are equal and similar to the three opposite.
Thus, the whole solid $CD$ is similar to the whole solid $AL$ [Def. 11.9] .
Thus, $AL$, similar, and similarly laid out, to the given parallelepipedal solid $CD$, has been described on the given straight lines $AB$.
(Which is) the very thing it was required to do.
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