If two straight lines are at right angles to the same plane, then the straight lines are parallel.

Let the two straight lines *AB* and *CD* be at right angles to the plane of reference.

I say that *AB* is parallel to *CD.*

Let them meet the plane of reference at the points *B* and *D.*

Join the straight line *BD.* Draw *DE* in the plane of reference at right angles to *BD,* and make *DE* equal to *AB.*

Now, since *AB* is at right angles to the plane of reference, it also makes right angles with all the straight lines which meet it and lie in the plane of reference.

But each of the straight lines *BD* and *BE* lies in the plane of reference and meets *AB,* therefore each of the angles *ABD* and *ABE* is right. For the same reason each of the angles *CDB* and *CDE* is also right.

And since *AB* equals *DE,* and *BD* is common, therefore the two sides *AB* and *BD* equal the two sides *ED* and *DB.* And they include right angles, therefore the base *AD* equals the base *BE*

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

But the angle *ABE* is right, therefore the angle *EDA* is also right. Therefore *ED* is at right angles to *DA.*

But it is also at right angles to each of the straight lines *BD* and *DC,* therefore *ED* is set up at right angles to the three straight lines *BD, DA,* and *DC* at their intersection. Therefore the three straight lines *BD, DA,* and *DC* lie in one plane.

But in whatever plane *DB* and *DA* lie, *AB* also lies, for every triangle lies in one plane.

Therefore the straight lines *AB, BD,* and *DC* lie in one plane. And each of the angles *ABD* and *BDC* is right, therefore *AB* is parallel to *CD.*

Therefore, *if two straight lines are at right angles to the same plane, then the straight lines are parallel.*

Q.E.D.

A converse of this proposition is XI.8.