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.
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"