Proof: By Euclid
(related to Proposition: Prop. 11.14: Planes Perpendicular to same Straight Line are Parallel)
- For let some straight line $AB$ be at right angles to each of the planes $CD$ and $EF$.
I say that the planes are parallel.
For, if not, being produced, they will meet.
- Let them have met.
- So they will make a straight line as a common section [Prop. 11.3].
- Let them have made $GH$.
- And let some random point $K$ have been taken on $GH$.
- And let $AK$ and $BK$ have been joined.
- And since $AB$ is at right angles to the plane $EF$, $AB$ is thus also at right angles to $BK$, which is a straight line in the produced plane $EF$ [Def. 11.3] .
- Thus, angle $ABK$ is a right angle.
- So, for the same (reasons), $BAK$ is also a right angle.
- So the (sum of the) two angles $ABK$ and $BAK$ in the triangle $ABK$ is equal to two right angles.
- The very thing is impossible [Prop. 1.17].
- Thus, planes $CD$ and $EF$, being produced, will not meet.
- planes $CD$ and $EF$ are thus parallel [Def. 11.8] .
- Thus, planes to which the same straight line is at right angles are parallel planes.
- (Which is) the very thing it was required to show.
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"