Consider a regular pentagon ABCDE with side length b.
Draw isosceles triangle XCD, such that the area of XCD is the same
as the area of the pentagon.
XC crosses AB at point Y.
XD crosses AE at point Z.
Show that the length of YZ equals 2b*cos(72).
[KD: This is the same as showing the distance YZ = BE - CD.]
What is the area of XCD outside the pentagon in terms of b?
For example, use b=4000.