4. In square $A B C D$, $M$ is the midpoint of $B C$. If $\overrightarrow{A C}=\lambda \overrightarrow{A M}+\mu \overrightarrow{B D}$, then $\lambda+\mu=$
A. $\frac{4}{3}$
B. $\frac{5}{3}$
C. $\frac{15}{8}$
D. 2
$\overrightarrow{A C}=\overrightarrow{A M}+\overrightarrow{M C}=\overrightarrow{A M}+\frac{1}{2} \overrightarrow{B C}$, since $\overrightarrow{A C}=\overrightarrow{B C}-\overrightarrow{B A}, \overrightarrow{B D}=\overrightarrow{B C}+\overrightarrow{B A}$
$\Rightarrow \overrightarrow{B C}=\frac{1}{2}(\overrightarrow{A C}+\overrightarrow{B D})$, thus $\overrightarrow{A C}=\overrightarrow{A M}+\frac{1}{4}(\overrightarrow{A C}+\overrightarrow{B D})$
$\Rightarrow \frac{3}{4} \overrightarrow{A C}=\overrightarrow{A M}+\frac{1}{4} \overrightarrow{B D} \Rightarrow \overrightarrow{A C}=\frac{4}{3} \overrightarrow{A M}+\frac{1}{3} \overrightarrow{B D} \Rightarrow \lambda+\mu=\frac{5}{3}$, so the answer is $B$.