Consider $\triangle A B C$ and $\triangle A_1 B_1 C_1$ in such a way that $\vec{A B}=\vec{A_1 B_1}$ and $M, N, M_1, N_1$ be the mid points of $A B, B C, A_1 B_1$ and $B_1 C_1$ respectively, then

$\vec{MM_1}=\vec{BB_1}$

$\vec{AB}=\vec{A_1 B_1} \quad \Rightarrow \vec{b}-\vec{a}=\vec{b}_1-\vec{a}_1$

$\Rightarrow \vec{b}-\vec{b}_1=\vec{a}-\vec{a}_1 \Rightarrow \vec{B_1 B}=\vec{A_1 A} \Rightarrow \vec{AA_1}=\vec{BB_1}$

$\Rightarrow \vec{NN_1}=\frac{\vec{b}_1+\vec{c}_1}{2}-\frac{\vec{b}+\vec{c}}{2} \Rightarrow \vec{NN_1}=\frac{\vec{b}_1+\vec{c}_1-\vec{b}-\vec{c}}{2} \Rightarrow 2 \vec{NN_1}=\vec{BB_1}+\vec{CC_1}$

$\Rightarrow \vec{MM_1}=\frac{\vec{b}_1-\vec{b}+\vec{a}_1-\vec{a}}{2} \Rightarrow 2 \vec{MM_1}=\vec{BB_1}+\vec{AA_1}=2 \vec{BB_1}=2 \vec{AA_1}$

$\Rightarrow \vec{MM_1}=\vec{BB_1}=\vec{AA_1}$

Hence (4) is correct answer.