If A, B, C, D be any four points and E and F be the middle points of AC and BD respectively, then $\vec{AB}+\vec{CB}+\vec{CD}+\vec{AD}$ is equal to

$4\vec{EF}$

Since F is the middle point of BD. Therefore,

$\vec{AB}+\vec{AD}=2\vec{AF}$   ...(i)

Similarly, we have

$\vec{CB}+\vec{CD}=2\vec{CF}$   ...(ii)

Adding (i) and (ii), we get

$\vec{AB}+\vec{AD}+\vec{CB}+\vec{CD}=2(\vec{AF}+\vec{CF})=-2(\vec{FA}+\vec{FC})$

$⇒\vec{AB}+\vec{AD}+\vec{CB}+\vec{CD}=-2(2\vec{FE})$

$⇒\vec{AB}+\vec{AD}+\vec{CB}+\vec{CD}=4\vec{EF}$  [∵ E is the mid-point of AC]