---
book: 11
number: 5
id: "XI.5"
kind: "theorem"
uses: ["[[book-11/proposition-3]]", "[[book-11/proposition-4]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.5"
license: "CC-BY-SA-4.0"
---

# XI.5

*If a straight line be set up at right angles to three straight lines which meet one another*, *at their common point of section*, *the three straight lines are in one plane.*

## Proof

For let a straight line *AB* be set up at right angles to the three straight lines *BC*, *BD*, *BE*, at their point of meeting at *B*; I say that *BC*, *BD*, *BE* are in one plane.

For suppose they are not, but, if possible, let *BD*, *BE* be in the plane of reference and *BC* in one more elevated; let the plane through *AB*, *BC* be produced; it will thus make, as common section in the plane of reference, a straight line. [[book-11/proposition-3|XI. 3]]

Let it make *BF*.

Therefore the three straight lines *AB*, *BC*, *BF* are in one plane, namely that drawn through *AB*, *BC*.

Now, since *AB* is at right angles to each of the straight lines *BD*, *BE*, therefore *AB* is also at right angles to the plane through *BD*, *BE*. [[book-11/proposition-4|XI. 4]]

But the plane through *BD*, *BE* is the plane of reference; therefore *AB* is at right angles to the plane of reference.

Thus *AB* will also make right angles with all the straight lines which meet it and are in the plane of reference. [[book-11/definitions#Definition 3|XI. Def. 3]]

But *BF* which is in the plane of reference meets it; therefore the angle *ABF* is right.

But, by hypothesis, the angle *ABC* is also right; therefore the angle *ABF* is equal to the angle *ABC*.

And they are in one plane: which is impossible.

Therefore the straight line *BC* is not in a more elevated plane; therefore the three straight lines *BC*, *BD*, *BE* are in one plane.

Therefore, if a straight line be set up at right angles to three straight lines, at their point of meeting, the three straight lines are in one plane. Q. E. D.