Planes through Parallel Pairs of Meeting Lines are Parallel

Proof

 * Euclid-XI-15.png

Let $AB$ and $BC$ be two straight lines which meet each other at $B$.

Let $DE$ and $EF$ be two straight lines which meet each other at $E$ such that:
 * $AB$ is parallel to $DE$
 * $BC$ is parallel to $EF$.

It is to be demonstrated that the plane holding $AB$ and $BC$ is parallel to the plane holding $DE$ and $EF$.

From :
 * let $BG$ be drawn from $B$ perpendicular to the plane holding $DE$ and $EF$.

From :
 * let $GH$ be drawn from $G$ parallel to $ED$

and
 * let $GK$ be drawn from $G$ parallel to $EF$.

We have that $BG$ is perpendicular to the plane holding $DE$ and $EF$.

Therefore from :
 * $BG$ is perpendicular to all the straight lines which meet it and are in the plane holding $DE$ and $EF$.

But each of $GH$ and $GK$ meets $BG$ and is in the plane holding $DE$ and $EF$.

Therefore each of $\angle BGH$ and $\angle BGK$ is a right angle.

From :
 * $BA$ is parallel to $GH$.

Therefore from :
 * $\angle GBA$ and $\angle BGH$ equal two right angles.

But $\angle BGH$ is a right angle.

Therefore $\angle GBA$ is a right angle.

Therefore $GB$ is perpendicular to $BA$.

For the same reason, $GB$ is also perpendicular to $BC$.

We have that the straight line $GB$ is set up perpendicular to the two straight lines $BA$ and $BC$ which cut one another.

Therefore from :
 * $GB$ is perpendicular to the plane holding $AB$ and $BC$.

But from :
 * the plane holding $AB$ and $BC$ is parallel to the plane holding $DE$ and $EF$.