Parallelepiped formed from Three Proportional Lines equal to Equilateral Parallelepiped with Equal Angles to it formed on Mean

Proof

 * Euclid-XI-36.png

Let $A, B, C$ be three straight lines in proportion:
 * $A : B = B : C$

Let a solid angle be constructed at $E$ contained by the plane angles $\angle DEG, \angle GEF, \angle FED$.

Let each of the straight lines $DE, GE, EF$ be made equal to $B$.

Let the parallelepiped $EK$ be completed.

Let the straight line $LM$ be made equal to $A$.

Let a solid angle be constructed at $L$ contained by the plane angles $\angle NLO, \angle OLM, \angle MLN$.

Let the straight line $LO$ be made equal to $B$.

Let the straight line $LN$ be made equal to $C$.

We have that:
 * $A : B = B : C$

while:
 * $A = LM$
 * $B = LO = ED = EF$
 * $C = LN$

Therefore:
 * $LM = EF = DE : LN$

Thus the sides about the equal angles $\angle NLM, \angle DEF$ are reciprocally proportional.

Therefore from :
 * the parallelogram $MN$ equals the parallelogram $DF$.

It follows from :
 * the perpendiculars from $G$ and $O$ to the planes containing $\angle NLM$ and $\angle DEF$ are equal.

Therefore the parallelepipeds $LH$ and $EK$ are of equal height.

So from :
 * $LH = EK$.

We have that:
 * $LH$ is the parallelepiped formed out of $A, B, C$
 * $EK$ is the parallelepiped formed out of $B$.

Hence the result.