Four Straight Lines are Proportional iff Similar Parallelepipeds formed on them are Proportional

Proof

 * Euclid-XI-37.png

Let $AB$, $CD$, $EF$ and $GH$ be four straight lines in proportion, so that:
 * $AB : CD = EF : GH$

Let similar and similarly situated parallelepipeds $KA$, $LC$, $ME$ and $NG$ be described on $AB$, $CD$, $EF$ and $GH$ respectively.

It is to be demonstrated that:
 * $KA : LC = ME : NG$

From :
 * $KA : LC = AB^3 : CD^3$

For the same reason:
 * $ME : NG = EF^3 : GH^3$

We have that:
 * $AB : CD = EF : GH$

Therefore:
 * $KA : LC = ME : NG$

Suppose that:
 * $KA : LC = ME : NG$

Then from :
 * $KA : LC = AB^3 : CD^3$

For the same reason:
 * $ME : NG = EF^3 : GH^3$

We have that:
 * $KA : LC = ME : NG$

Therefore:
 * $AB : CD = EF : GH$