證明：過E作GF⊥BC,交BC于F,交DC延長線于G
          ∵AB∥CD
          ∴EG⊥DG
          又∵E是BC的中點(diǎn)
          ∴GE=EF
          ∴S△DCE=DC*GE/2
              S△ABE=AB*EF/2
              S梯形ABCD=（DC+AB）*GF/2
          又∵S△ADE=S梯形ABCD-  S△DCE- S△ABE
          ∴S△ADE=（DC+AB）*GF/2-DC*GE/2-AB*EF/2
                          =（2DC*GE+2AB*GE-DC*GE-AB*GE）/2
                          =（DC*GE+AB*GE）/2
                          =[（DC+AB）*GE]/2
                          =[（DC+AB）*GF/2]/2
                          =S梯形ABCD/2
          即S△ADE=二分之一S梯形ABCD