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

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,-2.5){.} \psline{-}(-3,-2)(7,-2) \rput[a](7,-1.8){$\ell$} \pscircle[linecolor=blue](0,0){2.5} \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,-4.5){.} \psline{-}(-3,-2)(7,-2) \rput[a](7,-1.8){$\ell$} \pscircle(0,0){2.5} \pscircle[linecolor=blue](1.5,-2){2.5} \psdots(0,0)(1.5,-2)(4,-2) \rput[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,-4.5){.} \psline{-}(-3,-2)(7,-2) \rput[a](7,-1.8){$\ell$} \pscircle[linecolor=green](0,0){2.5} \pscircle(1.5,-2){2.5} \pscircle[linecolor=red](4,-2){2.5} \psline[linecolor=blue]{-}(-3,0)(7,0) \psdots(0,0)(1.5,-2)(4,-2)(2.5,0) \rput[a](0,0.2){$P$} \rput[a](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 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 link provided.