(1) The cross m in D stands for MD⊥AC, and the cross b in E stands for BE⊥AC.
∵∠a=∠a,∴rtδamd∽rtδabe，
∴BD/BE=AM/AB=2/3,∴BD=2/3BE，
∫sδAMN = 1/2 * MD * AN = 1/2 * 2/3BE * 1/2AC = 1/6 * BE * AC，
sδABC = 1/2 * BE * AC，
∴sδabc=3sδamn，
∴S quadrilateral bcnm = 2s δ AMN.
∴ s δ ABC/s quadrilateral BCNM=3/2.
(2) If the diameter is AF and BF and BC are connected in G, then ∠ ABF = 90,
∴BE=√(AF^2-AB^2)=2，
∴sδabf= 1/2*ab*bf=4，
sδabf= 1/2*af*bg,∴bg=4/√5。
∴BC=8/√5,AG=√(AB^2-BG^2)=8/√5
sδABC = 1/2BC * AG = 32/√5，
∴S quadrilateral bcnm = 2/3s δ ABC = 64 √ 5/ 15.