6. As shown in Figure 1, in the convex quadrilateral $A B C D$, $E$ and $F$ are the midpoints of sides $A B$ and $C D$ respectively. $A F$ and $D E$ intersect at point $G$, and $B F$ and $C E$ intersect at point $H$. The area of quadrilateral $E G F H$ is 10. Then the sum of the areas of $\triangle A D G$ and $\triangle B C H$ is $(\quad$.
(A) $\frac{20}{3}$
(B) 10
(C) 15
(D) 20
6. B.

Solution 1: Specialization.
From $A B / / B C$, we get
$$
S_{\triangle A X C}=S_{\triangle E C F}, S_{\triangle B H C}=S_{\triangle E H F}.
$$

Therefore, $S_{\triangle U D G}+S_{\triangle B M C}=S_{\triangle N C F}+S_{\triangle E H F}$
$$
=S_{\text {quadrilateral } E C F H}=10.
$$

Solution 2:
As shown in Figure 4, draw the heights
$A I, E J, B K$ from points $A, E, B$
to side $D C$, and connect
$A C, B D$.
Since $E$ is the midpoint of side $A B$, we have
$$
E J=\frac{1}{2}(A I+B K).
$$

Thus, $S_{\triangle D C E}=\frac{1}{2}\left(S_{\triangle A D C}+S_{\triangle B D C}\right)$.
Similarly, $S_{\triangle A B E}=\frac{1}{2}\left(S_{\triangle A A B}+S_{\triangle A B B}\right)$.
Therefore, $S_{\triangle E D C}+S_{\triangle F A B}$
$$
\begin{array}{l}
=\frac{1}{2}\left(S_{\triangle D C E}+S_{\triangle B C B}\right)+\frac{1}{2}\left(S_{\triangle D B B}+S_{\triangle C A B}\right) \\
=S_{\text {pentagon } A B C D}.
\end{array}
$$

As shown in Figure 4