6. In a convex quadrilateral $ABCD$, $AB \times CD$, $M$ is the midpoint of $AB$, $N$ is a point on $CD$, $AN$ intersects $DM$ at point $E$, and $CM$ intersects $BN$ at point $F$. If
$$
S_{\triangle ADE} + S_{\triangle BCF} = S_{\text{quadrilateral MENF}},
$$

then the value of $\frac{DN}{CD}$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) $\frac{2}{5}$
(D) $\frac{3}{5}$
6. A.

From the given equation, we know that
$$
S_{\triangle A D N}+S_{\triangle B C N}=S_{\triangle M D C} \text {. }
$$

As shown in Figure 3, draw perpendiculars from points $A$, $B$, and $M$ to $CD$, intersecting $CD$ at points $A'$, $B'$, and $M'$, respectively. Then
$$
\begin{array}{l}
A A' / / M M' / / B B' . \\
\text { Let } D N=x, \\
C N=y, \\
A A'=a, B B'=b .
\end{array}
$$

From the given information, $a \neq b$, and $M M'=\frac{1}{2}(a+b)$.
Thus, equation (1) becomes
$$
\frac{1}{2} a x+\frac{1}{2} b y=\frac{1}{2}(a+b) \cdot \frac{1}{2}(x+y),
$$

which simplifies to $(x-y)(a-b)=0$.
Therefore, $x-y=0 \Rightarrow x=y \Rightarrow \frac{D N}{C D}=\frac{1}{2}$.