---
book: 11
number: 16
id: "XI.16"
kind: "theorem"
uses: ["[[book-11/proposition-1]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.16"
license: "CC-BY-SA-4.0"
---

# XI.16

*If two parallel planes be cut by any plane*, *their common sections are parallel.*

## Proof

For let the two parallel planes *AB*, *CD* be cut by the plane *EFGH*, and let *EF*, *GH* be their common sections; I say that *EF* is parallel to *GH*.

For, if not, *EF*, *GH* will, when produced, meet either in the direction of *F*, *H* or of *E*, *G*.

Let them be produced, as in the direction of *F*, *H*, and let them, first, meet at *K*.

Now, since *EFK* is in the plane *AB*, therefore all the points on *EFK* are also in the plane *AB*. [[book-11/proposition-1|XI. 1]]

But *K* is one of the points on the straight line *EFK*; therefore *K* is in the plane *AB*.

For the same reason *K* is also in the plane *CD*; therefore the planes *AB*, *CD* will meet when produced.

But they do not meet, because they are, by hypothesis, parallel; therefore the straight lines *EF*, *GH* will not meet when produced in the direction of *F*, *H*.

Similarly we can prove that neither will the straight lines *EF*, *GH* meet when produced in the direction of *E*, *G*.

But straight lines which do not meet in either direction are parallel. [[book-1/definitions#Definition 23|I. Def. 23]]

Therefore *EF* is parallel to *GH*.

Therefore etc. Q. E. D.