5. Let the area of $\triangle A B C$ be $1, D$ be a point on side $A B$, and $\frac{A D}{A B}=\frac{1}{3}$. If a point $E$ is taken on side $A C$ such that the area of quadrilateral $D E C B$ is $\frac{3}{4}$, then the value of $\frac{C E}{E A}$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{1}{4}$
(D) $\frac{1}{5}$
5. (B).

As shown in Figure 3, connect $B E$,
$$
S_{\triangle A D E}=1-\frac{3}{4}=\frac{1}{4} \text {. }
$$

Let $\frac{C E}{A C}=x$, then
$$
S_{\triangle A B E}=1-x \text {. }
$$
$$
S_{\triangle A D E}=\frac{1-x}{3}=\frac{1}{4}, x=\frac{1}{4} \text {. Therefore, } \frac{C E}{E A}=\frac{1}{3} \text {. }
$$