Line Parallel to Perpendicular Line to Plane is Perpendicular to Same Plane

Theorem

In the words of Euclid:

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.

(The Elements: Book $\text{XI}$: Proposition $8$)

Proof

Let $AB$ and $CD$ be two parallel straight lines.

Let $AB$ be at right angles to the plane of reference.

Then $CD$ needs to be demonstrated to be also at right angles to the plane of reference.

Let $AB$ and $CD$ meet the plane of reference at $B$ and $D$.

Let $BD$ be joined.

Then by Proposition $7$ of Book $\text{XI} $: Line joining Points on Parallel Lines is in Same Plane:

$AB$, $CD$ and $BD$ are all in the same plane.

Let $DE$ be drawn in the plane of reference at right angles to $BD$ and equal to $AB$.

Let $BE$, $AE$ and $AD$ be joined.

We have that $AB$ is at right angles to the plane of reference.

Therefore from Book $\text{XI}$ Definition $3$: Line at Right Angles to Plane, $AB$ is at right angles to all the straight lines which meet it and are in the plane of reference.

Therefore each of $\angle ABD$ and $\angle ABE$ is a right angle.

We have that the straight line $BD$ falls on the parallels $AB$ and $CD$.

Therefore from Parallelism implies Supplementary Interior Angles $\angle ABD$ and $\angle CDB$ are equal to two right angles.

But $\angle ABD$ is a right angle.

Therefore $\angle CDB$ is also a right angle.

Therefore $CD$ is at right angles to $BD$.

We have that $AB = DE$, and that $BD$ is common.

The two sides $AB$ and $BD$ of $\triangle ABD$ are equal to the two sides $ED$ and $DB$ of $\triangle EBD$.

Also $\angle ABD = \angle EDB$, as they are both right angles.

Therefore $AD = BE$.

We have that:

$AB = DE$

and:

$BE = AD$

Thus the two sides $AB$ and $BE$ of $\triangle ABE$ are equal to the two sides $ED$ and $DA$ of $\triangle EAD$.

We have that $AE$ is common.

So:

$\angle ABE = \angle EDA$

But $\angle ABE$ is a right angle.

Therefore $\angle EDA$ is a right angle.

Therefore $ED$ is at right angles to $AD$.

But $ED$ is also at right angles to $DB$.

Therefore by Proposition $4$ of Book $\text{XI} $: Line Perpendicular to Two Intersecting Lines is Perpendicular to their Plane:

$ED$ is at right angles to the plane through $DA$ and $BD$.

But from Proposition $2$ of Book $\text{XI} $: Two Intersecting Straight Lines are in One Plane:

$DC$ is in the plane through $DA$ and $BD$ inasmuch as $AB$ and $BD$ are in the same plane which $AB$ and $BD$ are.

Also $DC$ is in the same plane which $AB$ and $BD$ are.

Therefore $ED$ is at right angles to $DC$.

So $CD$ 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 straight lines $DE$ and $DB$ which cut one another, at $D$.

So from Proposition $4$ of Book $\text{XI} $: Line Perpendicular to Two Intersecting Lines is Perpendicular to their Plane:

$CD$ is also at right angles to the plane through $DE$ and $DB$ are.

But the plane through $DE$ and $DB$ is the plane of reference.

Hence the result.

$\blacksquare$

Historical Note

This proof is Proposition $8$ of Book $\text{XI}$ of Euclid's The Elements.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{XI}$. Propositions