M,N分别是平行四边形ABCD对角线BD,AC上的点,AC,BD相交于点O.已知BM=1/3BO,ON=1/3OC.AB

MN = ON - OM
= 1/3OC - (OB+BM)
= 1/3OC - (OB+1/3BO)
= 1/3OC -2/3OB
= 1/3(OC-OB) + OB
= 1/3BC +OB
=1/3AD + OB
= (1/3)b + OB (1)
OB = DB/2
= (BC+CD)/2
= (AD +BA)/2 ( BC=AD,CD=BA)
= (b-a)/2 (2)
from(1) and (2)
MN = (1/3)b + OB
= (1/3)b + (b-a)/2
= (5/6)b -(1/2)a