---
book: 11
number: 36
id: "XI.36"
kind: "theorem"
uses: ["[[book-6/proposition-14]]", "[[book-11/proposition-35]]", "[[book-11/proposition-31]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.36"
license: "CC-BY-SA-4.0"
---

# XI.36

*If three straight lines be proportional*, *the parallelepipedal solid formed out of the three is equal to the parallelepipedal solid on the mean which is equilateral*, *but equiangular with the aforesaid solid.*

## Proof

Let *A*, *B*, *C* be three straight lines in proportion, so that, as *A* is to *B*, so is *B* to *C*; I say that the solid formed out of *A*, *B*, *C* is equal to the solid on *B* which is equilateral, but equiangular with the aforesaid solid.

Let there be set out the solid angle at *E* contained by the angles *DEG*, *GEF*, *FED*, let each of the straight lines *DE*, *GE*, *EF* be made equal to *B*, and let the parallelepipedal solid *EK* be completed, let *LM* be made equal to *A*, and on the straight line *LM*, and at the point *L* on it, let there be constructed a solid angle equal to the solid angle at *E*, namely that contained by *NLO*, *OLM*, *MLN*; let *LO* be made equal to *B*, and *LN* equal to *C*.

Now, since, as *A* is to *B*, so is *B* to *C*, while *A* is equal to *LM*, *B* to each of the straight lines *LO*, *ED*, and *C* to *LN*, therefore, as *LM* is to *EF*, so is *DE* to *LN*.

Thus the sides about the equal angles *NLM*, *DEF* are reciprocally proportional; therefore the parallelogram *MN* is equal to the parallelogram *DF*. [[book-6/proposition-14|VI. 14]]

And, since the angles *DEF*, *NLM* are two plane rectilineal angles, and on them the elevated straight lines *LO*, *EG* are set up which are equal to one another and contain equal angles with the original straight lines respectively, therefore the perpendiculars drawn from the points *G*, *O* to the planes through *NL*, *LM* and *DE*, *EF* are equal to one another; [[book-11/proposition-35|XI. 35]], Por. hence the solids *LH*, *EK* are of the same height.

But parallelepipedal solids on equal bases and of the same height are equal to one another; [[book-11/proposition-31|XI. 31]] therefore the solid *HL* is equal to the solid *EK*.

And *LH* is the solid formed out of *A*, *B*, *C*, and *EK* the solid on *B*; therefore the parallelepipedal solid formed out of *A*, *B*, *C* is equal to the solid on *B* which is equilateral, but equiangular with the aforesaid solid. Q. E. D.