Two Lines Perpendicular to Same Plane are Parallel

Proof

 * Euclid-XI-6.png

Let $AB$ and $CD$ be two straight lines at right angles to the plane of reference.

It is to be demonstrated that $AB$ is parallel to $CD$.

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

Let the straight line $BD$ be joined.

Let $DE$ be drawn in the plane of reference at right angles to $BD$ such that $DE = AB$.

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

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

So by :
 * $AB$ is at right angles to every straight line which meets it and is in the plane of reference.

But each of $BD$ and $BE$ is in the plane of reference and meets $AB$.

Therefore $\angle ABD$ and $\angle ABE$ are both right angles.

For the same reason $\angle CDB$ and $\angle CDE$ are both right angles.

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

Thus the two sides $AB$ and $BD$ of $\triangle ABD$ equal the two sides $BD$ and $BE$ of $\triangle EDB$.

The triangles $\triangle ABD$ and $\triangle EDB$ both include right angles.

Therefore by :
 * $AD = BE$

We have that:
 * $AB = DE$

and:
 * $AD = BE$

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

We also have that $AE$ is common.

So from :
 * $\angle ABE = \angle EDA$

But $\angle ABE$ is a right angle.

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

But $ED$ is also at right angles to the straight lines $BD$ and $DC$.

Therefore $ED$ is set up at right angles to the three straight lines $BD$, $DA$ and $DC$ at their meeting points.

Therefore from :
 * $BD$, $DA$ and $DC$ are in the same plane.

But from :
 * the triangle $DAB$ is in one plane.

Therefore $AB$ is in the same plane as $DB$ and $DA$.

Therefore the straight lines $AB$, $BD$ and $DC$ are in one plane.

Also, each of $\angle ABD$ and $\angle BDC$ is a right angle.

Therefore from Supplementary Interior Angles implies Parallel Lines:
 * $AB$ is parallel to $CD$.