Quandaries
and Queries 

I have been searching for
instructions on how to do this using a Thanks, 

Hi Mandy, The key to this construction is in the diagram below. DE is parallel to AC so the triangles ABC and DBE are similar. Thus ^{DB}/_{AB} = ^{EB}/_{CB}. Hence, for example, if DB is one third AB then EB is one third CB. The construction for dividing the line segment BC into three congruent segments is then as follows. Draw a ray through B at any convenient angle. With the compass point on B and any convenient length draw an arc intersecting the ray through B at P. Keeping the same compass setting, move the point to P and draw an arc intersecting the ray at Q. Again, with the same setting, move the point to Q and draw an arc, intersecting the ray at A. Join A to C. If you now construct a line through P that is parallel to AC and which intersects BC at R then BR is one third of BC. Can you construct a line through P that is parallel to AC? Penny 

