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[parent] compass and straightedge construction of parallel line (Algorithm)

Task. Construct the line parallel to a given line $ \ell$ and passing through a given point $ P$ which is not on $ \ell$.


\begin{pspicture}(-3,-2)(7,1) \rput[l](-3,-2){.} \psline{-}(-3,-2)(7,-2) \rput[a](7,-1.8){$\ell$} \psdot(0,0) \rput[a](0,0.25){$P$} \end{pspicture}

Solution.

  1. Draw a circle $ c_1$ with center $ P$ and intersecting $ \ell$ at two points, one of which is $ A$.

    \begin{pspicture}(-3,-2)(7,3) \rput[l](-3,-2){.} \rput[a](0,2.5){.} \rput[b](0,-... ...psdots(0,0)(1.5,-2) \rput[a](0,0.2){$P$} \rput[r](1.3,-1.8){$A$} \end{pspicture}
  2. Draw a second circle $ c_2$ with center $ A$ and the same radius $ r$ as $ c_1$. This circle also intersects $ \ell$ at two points, one of which is $ B$.

    \begin{pspicture}(-3,-5)(7,3) \rput[l](-3,-2){.} \rput[a](0,2.5){.} \rput[b](1.5... ...t[a](0,0.2){$P$} \rput[a](1.3,-1.8){$A$} \rput[a](4.2,-2.2){$B$} \end{pspicture}
  3. Draw a third circle $ c_3$ with center $ B$ and radius $ r$. Let $ C$ be the intersection point of $ c_3$ (drawn below in red) with $ c_1$ (drawn below in green) which lies on the same side of $ \ell$ as $ P$ does. The line $ PC$ (drawn below in blue) is the required parallel to $ \ell$.

    \begin{pspicture}(-3,-5)(7,3) \rput[l](-3,-2){.} \rput[a](0,2.5){.} \rput[b](1.5... ...](1.3,-1.8){$A$} \rput[a](4.2,-2.2){$B$} \rput[a](2.7,-0.2){$C$} \end{pspicture}

Note 1. The construction is based on the fact that the quadrilateral $ PABC$ is a parallelogram. In fact, $ PABC$ is a rhombus. The reasoning is as follows:

  • The green circle shows that $ \overline{PC}$ and $ \overline{PA}$ are congruent.
  • The black circle shows that $ \overline{PA}$ and $ \overline{AB}$ are congruent.
  • The red circle shows that $ \overline{AB}$ and $ \overline{BC}$ are congruent.
  • Since $ PABC$ is a quadrilateral with all sides congruent, it is a rhombus (and therefore a parallelogram).
Note 2. It is clear that the construction only needs the compass, not a straightedge: In determining the point $ C$, the straightedge is totally superfluous, and the points $ P$ and $ C$ determine the desired line (which thus is not necessary to actually draw!). It may be proved that all constructions with compass and straightedge are possible using only the compass.

Note 3. The construction of the parallel uses the fact that the endpoints of two congruent chords (red) in a circle determine two parallel chords:


\begin{pspicture}(-5,0)(5,5) \rput[l](-5.5,2){.} \rput[a](0,5){.} \rput[b](-0.05... ...dashed](0,0)(4.89,1) \rput[r](5.8,2){$l$} \psarc(0,0){5}{5}{170} \end{pspicture}

If you are interested in seeing the rules for compass and straightedge constructions, click on the link provided.



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"compass and straightedge construction of parallel line" is owned by pahio. [ full author list (2) ]
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See Also: parallel postulate, $n$-section of line segment with compass and straightedge

Other names:  construction of parallel, construction of parallel line
Keywords:  Euclidean geometry

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Cross-references: compass and straightedge constructions, chords, endpoints, necessary, straightedge, compass, clear, congruent, rhombus, parallelogram, quadrilateral, side, intersects, radius, circle, point, passing through, parallel, line
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This is version 20 of compass and straightedge construction of parallel line, born on 2007-06-03, modified 2007-10-22.
Object id is 9506, canonical name is CompassAndStraightedgeConstructionOfParallelLine.
Accessed 1851 times total.

Classification:
AMS MSC: 51-00 (Geometry :: General reference works )
 51M15 (Geometry :: Real and complex geometry :: Geometric constructions)
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