Here's one proposed solution. Construct an isosceles triangle with base BC equal to 1 and height AD equal to 2.
Trisect the base BC at points E and F, that is, make BE = EF = FC, each of length 1/3. Then connect E and F to the vertex A.
Let G be the midpoint of the side AB. Construct an isosceles triangle BGH with H a point on the other side AC, and BG = BH.
Let GH intersect AE at L, and GH intersect AF at J. Also, let BH intersect AE at O, and BH intersect AF at N.
Then draw lines parallel to GH through N and O, and let them intersect AB at K and I, respectively. Also draw lines KL and IJ parallel to BH.
The three lines IJ, AE, and KN meet at a point. (They look like they do,anyway.)
It was proved in the 19th century that a cube can not be duplicated with the Euclidean tools of straightedge and compass.
This page: June, 2001
David E. Joyce
Department of Mathematics and Computer Science
Worcester, MA 01610