I got the final piece in the problem regarding figure 31 in the text.  It was pretty easy to get GA = AB = EG = EF.  (I hope that I got all of the letters right, I am not looking at the figure as I write this).  Anyways, in a rectangle the diagonals are congruent and bisecr each other.  So if you complete the rectangle that contains the triangle EBF with EF as the hypotenuse it is easy to conclude that the point G, which is the midpoint of EF is equidistant from the vertices of the traingle EBF.

On another note.  What problems do you want to work on on Monday night?  Noone has posted any suggestions.  Should I?  I would rather you pick your favorites.

               --CFW