如图,矩形ABCD中,AB=2,BC=3,对角线AC的垂直平分线分别交AD丶BC于点F

求EF的长吗?连接 CE,∵EF垂直平分AC,∴AE=CE,高DE=X,则AE=CE=3-X,在RTΔCDE中,CE^2=DE^2+CD^2,∴(3-X)^2=X^2+4,X=5/6,∴AE=CF=3-5/6=13/5,过E作EH⊥BC于H,则FH=13/6-5/6=4/3,∴EF=√(FH^2+EH^2)=√34/3....