---
book: 11
number: 4
id: "XI.4"
kind: "theorem"
uses: ["[[book-1/proposition-15]]", "[[book-1/proposition-4]]", "[[book-1/proposition-26]]", "[[book-1/proposition-8]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.4"
license: "CC-BY-SA-4.0"
---

# XI.4

*If a straight line be set up at right angles to two straight lines which cut one another*, *at their common point of section*, *it will also be at right angles to the plane through them.*

## Proof

For let a straight line *EF* be set up at right angles to the two straight lines *AB*, *CD*, which cut one another at the point *E*, from *E*; I say that *EF* is also at right angles to the plane through *AB*, *CD*.

For let *AE*, *EB*, *CE*, *ED* be cut off equal to one another, and let any straight line *GEH* be drawn across through *E*, at random; let *AD*, *CB* be joined, and further let *FA*, *FG*, *FD*, *FC*, *FH*, *FB* be joined from the point *F* taken at random lton *EF*gt.

Now, since the two straight lines *AE*, *ED* are equal to the two straight lines *CE*, *EB*, and contain equal angles, [[book-1/proposition-15|I. 15]] therefore the base *AD* is equal to the base *CB*, and the triangle *AED* will be equal to the triangle *CEB*; [[book-1/proposition-4|I. 4]] so that the angle *DAE* is also equal to the angle *EBC*.

But the angle *AEG* is also equal to the angle *BEH*; [[book-1/proposition-15|I. 15]] therefore *AGE*, *BEH* are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely that adjacent to the equal angles, that is to say, *AE* to *EB*; therefore they will also have the remaining sides equal to the remaining sides. [[book-1/proposition-26|I. 26]]

Therefore *GE* is equal to *EH*, and *AG* to *BH*.

And, since *AE* is equal to *EB*, while *FE* is common and at right angles, therefore the base *FA* is equal to the base *FB*. [[book-1/proposition-4|I. 4]]

For the same reason *FC* is also equal to *FD*.

And, since *AD* is equal to *CB*, and *FA* is also equal to *FB*, the two sides *FA*, *AD* are equal to the two sides *FB*, *BC* respectively; and the base *FD* was proved equal to the base *FC*; therefore the angle *FAD* is also equal to the angle *FBC*. [[book-1/proposition-8|I. 8]]

And since, again, *AG* was proved equal to *BH*, and further *FA* also equal to *FB*, the two sides *FA*, *AG* are equal to the two sides *FB*, *BH*.

And the angle *FAG* was proved equal to the angle *FBH*; therefore the base *FG* is equal to the base *FH*. [[book-1/proposition-4|I. 4]]

Now since, again, *GE* was proved equal to *EH*, and *EF* is common, the two sides *GE*, *EF* are equal to the two sides *HE*, *EF*; and the base *FG* is equal to the base *FH*; therefore the angle *GEF* is equal to the angle *HEF*. [[book-1/proposition-8|I. 8]]

Therefore each of the angles *GEF*, *HEF* is right.

Therefore *FE* is at right angles to *GH* drawn at random through *E*.

Similarly we can prove that *FE* will also make right angles with all the straight lines which meet it and are in the plane of reference.

But a straight line is at right angles to a plane when it makes right angles with all the straight lines which meet it and are in that same plane; [[book-11/definitions#Definition 3|XI. Def. 3]] therefore *FE* is at right angles to the plane of reference.

But the plane of reference is the plane through the straight lines *AB*, *CD*.

Therefore *FE* is at right angles to the plane through *AB*, *CD*.

Therefore etc. Q. E. D.