(1) because ∠DEF=∠ABC=60? , so DE∑AB, AD/BE=AM/BM=√3/ 1.

(2)① Connect AM and DM, let AB=a, DE=b,? ADM and? In the BEM, EMB=∠DMA=α,

AM/BM=DM/EM=√3/ 1, so these two triangles are almost the same, so AD/BE=√3/ 1.

②S=S(ABMD)-S(BME)-S (Mediterranean)

=S (anti-ballistic missile) +S(AMD)-S(BME)-S (Mediterranean)

= 1/2 * BM * AM+ 1/2 * AM * MH- 1/2 * BM * EM * sinα- 1/2 * ME * MD

= 1/2 * 3 * 3√3+ 1/2 * 3√3(x-3)- 1/2 * 3 * 1 * sinα- 1/2 * 1 *√3

=√3/2 *(3X-√3 sinα- 1)(3≤x≤3+√3)