将矩形ABCD沿直线BD折叠,使点C落在F处,BF交AD于E,AD=8,AB=4,求△BED的面积

1) ∠bae=∠dfe=90(矩形角) ab=dc=fd(对折) ∠abe=∠fde(因∠aeb=∠fed)△aeb≌△fde2) ae=fe be=de ae+de=8(bc)=ae+be3) △abe中,ae^2+ab^2=ae^2+4^2=be^2 be=5 ae=34) S△bed=1/2 *de*ab=1/2 *4*5=10