◀ ▲ ▶Branches / Geometry / Elements-euclid / Book-11-elementary-stereometry / Proof: By Euclid

Proof: By Euclid

(related to Proposition: Prop. 11.15: Planes through Parallel Pairs of Meeting Lines are Parallel)

• For let the two straight lines joined to one another, $AB$ and $BC$, be parallel to the two straight lines joined to one another, $DE$ and $EF$ (respectively), not being in the same plane.
• I say that the planes through $AB$, $BC$ and $DE$, $EF$ will not meet one another (when) produced.

• For let $BG$ have been drawn from point $B$ perpendicular to the plane through $DE$ and $EF$ [Prop. 11.11], and let it meet the plane at point $G$.
• And let $GH$ have been drawn through $G$ parallel to $ED$, and $GK$ (parallel) to $EF$ [Prop. 1.31].
• And since $BG$ is at right angles to the plane through $DE$ and $EF$, it will thus also make right angles with all of the straight lines joined to it, which are also in the plane through $DE$ and $EF$ [Def. 11.3] .
• And each of $GH$ and $GK$, which are in the plane through $DE$ and $EF$, are joined to it.
• Thus, each of the angles $BGH$ and $BGK$ are right angles.
• And since $BA$ is parallel to $GH$ [Prop. 11.9], the (sum of the) angles $GBA$ and $BGH$ is equal to two right angles [Prop. 1.29].
• And $BGH$ (is) a right angle.
• $GBA$ (is) thus also a right angle.
• Thus, $GB$ is at right angles to $BA$.
• So, for the same (reasons), $GB$ is also at right angles to $BC$.
• Therefore, since the straight line $GB$ has been set up at right angles to two straight lines, $BA$ and $BC$, cutting one another, $GB$ is thus at right angles to the plane through $BA$ and $BC$ [Prop. 11.4].
• So, for the same (reasons), $BG$ is also at right angles to the plane through $GH$ and $GK$.
• And the plane through $GH$ and $GK$ is the (plane) through $DE$ and $EF$.
• And it was also shown that $GB$ is at right angles to the plane through $AB$ and $BC$. And planes to which the same straight line is at right angles are parallel planes [Prop. 11.14].
• Thus, the plane through $AB$ and $BC$ is parallel to the (plane) through $DE$ and $EF$.
• Thus, if two straight lines joined to one another are parallel (respectively) to two straight lines joined to one another, which are not in the same plane, then the planes through them are parallel (to one another).
• (Which is) the very thing it was required to show.
  ∎

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References

Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"