Proof: By Euclid
(related to Proposition: Prop. 11.05: Three Intersecting Lines Perpendicular to Another Line are in One Plane)
- For let some straight line $AB$ have been set up at right angles to three straight lines $BC$, $BD$, and $BE$, at the (common) point of section $B$.
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I say that $BC$, $BD$, and $BE$ are in one plane.
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For (if) not, and if possible, let $BD$ and $BE$ be in the reference plane, and $BC$ in a more elevated (plane).
- And let the plane through $AB$ and $BC$ have been produced.
- So it will make a straight line as a common section with the reference plane [Def. 11.3] .
- Let it make $BF$.
- Thus, the three straight lines $AB$, $BC$, and $BF$ are in one plane - (namely), that drawn through $AB$ and $BC$.
- And since $AB$ is at right angles to each of $BD$ and $BE$, $AB$ is thus also at right angles to the plane (passing) through $BD$ and $BE$ [Prop. 11.4].
- And the plane (passing) through $BD$ and $BE$ is the reference plane.
- Thus, $AB$ is at "right angles to the reference plane":bookofproofs$2212.
- Hence, $AB$ will also make right angles with all straight lines joined to it which are also in the reference plane [Def. 11.3] .
- And $BF$, which is in the reference plane, is joined to it.
- Thus, the angle $ABF$ is a right angle.
- And $ABC$ was also assumed to be a right angle.
- Thus, angle $ABF$ (is) equal to $ABC$.
- And they are in one plane.
- The very thing is impossible.
- Thus, $BC$ is not in a more elevated plane.
- Thus, the three straight lines $BC$, $BD$, and $BE$ are in one plane.
- Thus, if a straight line is set up at right angles to three straight lines cutting one another, at the (common) point of section, then the three straight lines are in one plane.
- (Which is) the very thing it was required to show.
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References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"
