Question

如图,E F G H分别在菱形ABCD的四条边上 BE=BF=DG=DH,连接EF,FG,GH,E

Answer 1

连结AC、BD,则EF//=1/2AC、GH//=1/2AC=>四边形EFGH为平行四边形

当四边形ABCD的对角线应满足：AC⊥ BD、AC=BD时,四边形EFGH为正方形

AC⊥ BD,EF//AC,GF//BD=>∠ EFG=90°=>平行四边形EFGH为j矩形

EF=1/2AC,GF1/2BD,AC=BD=>EF=GF=>四边形EFGH为正方形