∴ vector AC=a+b, 3|a|=2|b|,

∴OA=μ(AB+2AC)=μ(3a+2b)，

On AD, the intercept AG=AB, and let the midpoint of BG be m, then

AM=( 1/2)(a+2b/3)，

OA=(3μ/2)AM，

Let the midpoint of AB and CD be e and f respectively,

OE=λOF from OA+OB=λ(OC+OD)，

∴O is the intersection point m of the straight line AM and EF.

∴λ= 1/2.