(2) If F is FH⊥BC in H, FH=CH= radical number 3*BH, BH+CH= 1, so FH=(3- radical number 3)/2, so CF=(3- radical number 3)/ radical number 2, AF= (radical number 3-60). Similarly, S△CGD=BO*AF/2= (root number 3- 1)/4. S△CEB= radical number 3/4, so S(AFEGD)= 1-2* (radical number 3- 1)/4- radical number 3/4=(6-3 radical number 3)/4.