Question

如图，将矩形ABCD沿着直线BD折叠,使点C落c处,Bc交AD于E,AD=8,AB=4,求三角形BED的面积？

Answer 1

解：有对称性可知∠DBC=∠C'BD

       又有AD//BC,∴∠CBD=∠ADB

         ∴EB=ED,

    在RT△BAE中有

          4²+AE²=(8-AE)²

   解得：AE=3，

    ∴  ED=5

      ∴  S△BED=ED*AB/2=5*4/2=10

Answer 2

由题意得：BC‘=BC=AD，易证AE=EC'
AE+ED=8 (1)
4²+AE²=(8-AE)² (2)
解得：AE=3，ED=5
S△BED=ED*AB/2=5*4/2=10

Answer 3

RT△BCD≌RT△BDC'
DC=DC'=AB=4
∠A=∠C'=90º
∠AEB=∠C'ED
∴RT△BAE≌RT△DC'E
∴AE=C'E