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

Proof

 * Euclid-XI-8.png

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 :
 * $AB$, $CD$ 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, $AB$ 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 :
 * $ED$ is at right angles to the plane through $DA$ and $BD$.

But from :
 * $DC$ 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 :
 * $CD$ 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.