compass and straightedge construction of parallel line

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

Solution.

1. Draw a circle $c_1$ with center (http://planetmath.org/Center8) $P$ and intersecting $\mathrm{\ell }$ at two points, one of which is $A$.

2. Draw a second circle $c_2$ with center $A$ and the same radius $r$ as $c_1$. This circle also intersects $\mathrm{\ell }$ at two points, one of which is $B$.

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 $\mathrm{\ell }$ as $P$ does.  The line $PC$ (drawn below in blue) is the required parallel to $\mathrm{\ell }$.

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

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