---
book: 11
number: 8
id: "XI.8"
kind: "theorem"
uses: ["[[book-11/proposition-7]]", "[[book-1/proposition-29]]", "[[book-11/proposition-4]]", "[[book-11/proposition-2]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.8"
license: "CC-BY-SA-4.0"
---

# XI.8

*If two straight lines be parallel*, *and one of them be at right angles to any plane*, *the remaining one will also be at right angles to the same plane.*

## Proof

Let *AB*, *CD* be two parallel straight lines, and let one of them, *AB*, be at right angles to the plane of reference; I say that the remaining one, *CD*, will also be at right angles to the same plane.

For let *AB*, *CD* meet the plane of reference at the points *B*, *D*, and let *BD* be joined; therefore *AB*, *CD*, *BD* are in one plane. [[book-11/proposition-7|XI. 7]]

Let *DE* be drawn, in the plane of reference, at right angles to *BD*, let *DE* be made equal to *AB*, and let *BE*, *AE*, *AD* be joined.

Now, since *AB* is at right angles to the plane of reference, therefore *AB* is also at right angles to all the straight lines which meet it and are in the plane of reference; [[book-11/definitions#Definition 3|XI. Def. 3]] therefore each of the angles *ABD*, *ABE* is right.

And, since the straight line *BD* has fallen on the parallels *AB*, *CD*, therefore the angles *ABD*, *CDB* are equal to two right angles. [[book-1/proposition-29|I. 29]]

But the angle *ABD* is right; therefore the angle *CDB* is also right; therefore *CD* is at right angles to *BD*.

And, since *AB* is equal to *DE*, and *BD* is common, the two sides *AB*, *BD* are equal to the two sides *ED*, *DB*; and the angle *ABD* is equal to the angle *EDB*, for each is right; therefore the base *AD* is equal to the base *BE*.

And, since *AB* is equal to *DE*, and *BE* to *AD*, the two sides *AB*, *BE* are equal to the two sides *ED*, *DA* respectively, and *AE* is their common base; therefore the angle *ABE* is equal to the angle *EDA*.

But the angle *ABE* is right; therefore the angle *EDA* is also right; therefore *ED* is at right angles to *AD*.

But it is also at right angles to *DB*; therefore *ED* is also at right angles to the plane through *BD*, *DA*. [[book-11/proposition-4|XI. 4]]

Therefore *ED* will also make right angles with all the straight lines which meet it and are in the plane through *BD*, *DA*.

But *DC* is in the plane through *BD*, *DA*, inasmuch as *AB*, *BD* are in the plane through *BD*, *DA*, [[book-11/proposition-2|XI. 2]] and *DC* is also in the plane in which *AB*, *BD* are.

Therefore *ED* is at right angles to *DC*, so that *CD* is also at right angles to *DE*.

But *CD* is also at right angles to *BD*.

Therefore *CD* is set up at right angles to the two straight lines *DE*, *DB* which cut one another, from the point of section at *D*; so that *CD* is also at right angles to the plane through *DE*, *DB*. [[book-11/proposition-4|XI. 4]]

But the plane through *DE*, *DB* is the plane of reference; therefore *CD* is at right angles to the plane of reference.

Therefore etc. Q. E. D.