I have it ...the 6th solution
1) a square with 4 sides and two diagonals =6 lines
2) would be a diamond shape created by two equilateral triangles; 5 sides equal and one long diagonal
3) An equilateral triangle ,3 equal sides ; The 4th point being in the center equidistant from the other 3 points providing 3 other lines of equal length but obviously shorter
4) an equilateral triangle is 3 lines connecting points "A" ,"B" and "C", Assume "A" is the vertex and "B" and "C" are the base points. A 4th point "X"gives us a line the same length as a sides of the triangle ,and directly off of vertex point "A" away from the triangle , but perpendicular to the opposing side "B"C"; This arrangement gives us 4 lines of equal length . Now the lines connecting the 2 base points "B"and "C", and point "X" ,are of equal length and longer than the other 4 lines
5) an equilateral triangle is 3 lines connecting points "A" ,"B" and "C", Assume "A" is the vertex and "B" and "C" are the base points. A 4th point "X"gives us a line the same length as a sides of the triangle ,and directly off of vertex point "A",but downward bisecting line BC = 4 lines of equal length; 2 short lines BX and CX
6) 4 points A,B,C,D, lay on two intersecting circles. point A is the center of one circle C and D lay on this circle. D is the center of the 2nd circle with A and B laying on the circle. The 4 points create a trapezoid in which the distance BC , is the same length as chords AB and CD (that is 3 short lengths ) The diagonals AC ,AD and BD are equal all being equal to the radius.
That should be all of them ????
6) 4 points A,B,C,D, lay on two intersecting circles having the same radius. point A is the center of one circle C and D lay on this circle. D is the center of the 2nd circle with A and B laying on the circle. The 4 points create a trapezoid in which the distance BC , is the same length as chords AB and CD (that is 3 short lengths ) The diagonals AC ,AD and BD are equal all being equal to the radius.
Actually I think this goes back a long way to the old Saturday evening post. The military had made a study of all of the planes that had made it back from air raids. Cataloged every bullet hole that had pierced the hull of these airplanes . There were suggestions that these areas should be armor plated . Someone else spoke up making the statement, that in fact these were the planes that made it back meaning these areas were less damaging to the plane ability to fly ,and the armor plating could best be used else where
Donald Cartmill's post in Line segments of two types was marked as the answer