梯形中位线定理的证明课件

EF=?（BC+AD）证明：连结AF并延长，BACEF三角形中位线定理三角形的中位线平行于第三边，AD//BC，EF=?BCM梯形中位线定理梯形的中位线平行于两底，DF=FC求证：EF//BC，AE=EB，交BC的延长线于点M∵DF=FC，∠D=∠3，并且等于它的一半即EF//BC，∠1=∠2，AD=CM又AE=EB∴EF是?ABM的中位线∴EF//BC，并且等于两底和的一半BCAEDF已知：梯形ABCD中，∴?ADF≌?MCF∴AF=MF，EF=?BM∵BM=BC+CM=BC+AD∴EF=?（BC+AD）1231BACDEFGBACDEFNMBACDEFMNBACDEFGBACDEFGBACDEFNMBACDEFGBACDEFMNBACDEFGBACDEFGBACDEFGBACDEFBACDEFNMMNBACDEFGBACDEFBACDEFNMGBACDEFGBACDEFBACDEFNMMNBACDEFGBACDEFBACDEFMNGBACDEFGBACDEFBACDEFNMGBACDEFGBACDEFBACDEFBACDEFNMMNGBACDEFGBACDEFBACDEFMNGBACDEFG，