Proof: By Euclid

Let $AB$, $CD$, $EF$, and $GH$, be four proportional straight lines, (such that) as $AB$ (is) to $CD$, so $EF$ (is) to $GH$.
And let the similar, and similarly laid out, parallelepiped solids $KA$, $LC$, $ME$ and $NG$ have been described on $AB$, $CD$, $EF$, and $GH$ (respectively).
I say that as $KA$ is to $LC$, so $ME$ (is) to $NG$.

fig37e

For since the parallelepipedal solid $KA$ is similar to $LC$, $KA$ thus has to $LC$ the cubed ratio that $AB$ (has) to $CD$ [Prop. 11.33].
So, for the same (reasons), $ME$ also has to $NG$ the cubed ratio that $EF$ (has) to $GH$ [Prop. 11.33].
And since as $AB$ is to $CD$, so $EF$ (is) to $GH$, thus, also, as $AK$ (is) to $LC$, so $ME$ (is) to $NG$.
And so let solid $AK$ be to solid $LC$, as solid $ME$ (is) to $NG$.
I say that as straight line $AB$ is to $CD$, so $EF$ (is) to $GH$.
For, again, since $KA$ has to $LC$ the cubed ratio that $AB$ (has) to $CD$ [Prop. 11.33], and $ME$ also has to $NG$ the cubed ratio that $EF$ (has) to $GH$ [Prop. 11.33], and as $KA$ is to $LC$, so $ME$ (is) to $NG$, thus, also, as $AB$ (is) to $CD$, so $EF$ (is) to $GH$.
Thus, if four straight lines are proportional, and so on of the proposition.
(Which is) the very thing it was required to show.
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