---
book: 11
number: 19
id: "XI.19"
kind: "theorem"
uses: ["[[book-11/proposition-13]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.19"
license: "CC-BY-SA-4.0"
---

# XI.19

*If two planes which cut one another be at right angles to any plane*, *their common section will also be at right angles to the same plane.*

## Proof

For let the two planes *AB*, *BC* be at right angles to the plane of reference, and let *BD* be their common section; I say that *BD* is at right angles to the plane of reference.

For suppose it is not, and from the point *D* let *DE* be drawn in the plane *AB* at right angles to the straight line *AD*, and *DF* in the plane *BC* at right angles to *CD*.

Now, since the plane *AB* is at right angles to the plane of reference, and *DE* has been drawn in the plane *AB* at right angles to *AD*, their common section, therefore *DE* is at right angles to the plane of reference. [[book-11/definitions#Definition 4|XI. Def. 4]]

Similarly we can prove that *DF* is also at right angles to the plane of reference.

Therefore from the same point *D* two straight lines have been set up at right angles to the plane of reference on the same side: which is impossible. [[book-11/proposition-13|XI. 13]]

Therefore no straight line except the common section *DB* of the planes *AB*, *BC* can be set up from the point *D* at right angles to the plane of reference.

Therefore etc. Q. E. D.