This time angle BED looks obtuse. Thus, there exists, for continuity reasons, a plane that passes through FC and gives a rectangular section. Thus one only has to find a point H on DF such that EH is perpendicular to EB . H lays in a plane p that passes through E and is perpendicular to EB . One draws the perpendiculars EG to DA , then GH to AD .
H is the sought point.
One just have to cut through B , E , H to get a rectangular section. Hutchinson Educational 1960. Epitsuk fel a matematikat. Universitatea din Timisoara, 1994. Three questions challenging the tradition of solid geometry teaching.