Proof: By Euclid

Let $AB$ and $CD$ be two parallel straight lines, and let one of them, $AB$, be at right angles to a reference plane.
I say that the remaining (one), $CD$, will also be at right angles to the same plane.
For let $AB$ and $CD$ meet the reference plane at points $B$ and $D$ (respectively).
And let $BD$ have been joined.
$AB$, $CD$, and $BD$ are thus in one plane [Prop. 11.7].
Let $DE$ have been drawn at right angles to $BD$ in the reference plane, and let $DE$ be made equal to $AB$, and let $BE$, $AE$, and $AD$ have been joined.
And since $AB$ is at "right angles to the reference plane":bookofproofs$2212, $AB$ is thus also at right angles to all of the straight lines joined to it which are in the reference plane [Def. 11.3] .
Thus, the angles $ABD$ and $ABE$ [are] each right angles.
And since the straight line $BD$ has met the parallel (straight lines) $AB$ and $CD$, the (sum of the) angles $ABD$ and $CDB$ is thus equal to two right angles [Prop. 1.29].
And $ABD$ (is) a right angle.
Thus, $CDB$ (is) also a right angle.
$CD$ is thus at right angles to $BD$.
And since $AB$ is equal to $DE$, and $BD$ (is) common, the two (straight lines) $AB$ and $BD$ are equal to the two (straight lines) $ED$ and $DB$ (respectively).
And angle $ABD$ (is) equal to angle $EDB$.
For each (is) a right angle.
Thus, the base $AD$ (is) equal to the base $BE$ [Prop. 1.4].
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 their base $AE$ is common.
Thus, angle $ABE$ is equal to angle $EDA$ [Prop. 1.8].
And $ABE$ (is) a right angle.
$EDA$ (is) thus also a right angle.
Thus, $ED$ is at right angles to $AD$.
And it is also at right angles to $DB$.
Thus, $ED$ is also at right angles to the plane through $BD$ and $DA$ [Prop. 11.4].
And $ED$ will thus make right angles with all of the straight lines joined to it which are also in the plane through $BDA$.
And $DC$ is in the plane through $BDA$, inasmuch as $AB$ and $BD$ are in the plane through $BDA$ [Prop. 11.2], and in which(ever plane) $AB$ and $BD$ (are found), $DC$ is also (found).
Thus, $ED$ is at right angles to $DC$.
Hence, $CD$ is also at right angles to $DE$.
And $CD$ is also at right angles to $BD$.
Thus, $CD$ is standing at right angles to two straight lines, $DE$ and $DB$, which meet one another, at the (point) of section, $D$.
Hence, $CD$ is also at right angles to the plane through $DE$ and $DB$ [Prop. 11.4].
And the plane through $DE$ and $DB$ is the reference (plane).
$CD$ is thus at "right angles to the reference plane":bookofproofs$2212.
Thus, if two straight lines are parallel, and one of them is at right angles to some plane, then the remaining (one) will also be at right angles to the same plane.
(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"

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Proof: By Euclid

References

Adapted from (subject to copyright, with kind permission)