在平行四边形ABCD中,EF过对角线的交点O,AB=4,AD=3四边形BCEF的周长=9.6求OF的长

AB∥CD,所以∠FAO=∠OCE,∠AFO=∠CEO,又∠AOF=∠EOC,△AFO∽△CEO
则AO/OC=AF/CE=OF/OE=1/1 所以AF=CE,OF=OE
四边形BCEF的周长=9.6
=BC+CE+OE+OF+BF=BC+AF+OF+OF+BF
=BC+AB+2OF=AD+AB+2OF=3+4+2OF
2OF=9.6-7
OF=1.3