Problem 868

【Analysis】Draw $M Q$ parallel to $B C$ intersecting $D F$ at $Q$, and draw $E P$ parallel to $A B$ intersecting $B M$ at $P$. Using the proportional relationships between line segments, find the ratio of the areas of the triangles, and finally determine the area of the shaded part.

【Solution】According to the analysis, as shown in the figure, draw $M Q$ parallel to $B C$ intersecting $D F$ at $Q$, and draw $E P$ parallel to $A B$ intersecting $B M$ at $P$,
$\because M$ is the midpoint of $C D$, so $Q M: P C=1: 2, \therefore Q M: B F=1: 4$, so $G M: G B=1: 4$, $\therefore B G: B M=4: 5$; also, since $B F: B C=2: 3, S_{\triangle \mathrm{BFG}}=\frac{4}{5} \times \frac{2}{3} \mathrm{~S}_{\triangle \mathrm{ABC}}=\frac{2}{15}$;
$\because E$ is the trisection point on side $B C$, so $E P: C M=1: 3, \therefore E P: A B=1: 6$,
$\therefore B H: H P=6: 1, \therefore B H: H M=6: 15=2: 5, B H: B G=2: 7$,
also $\because G M: \quad G B=1: 4, \therefore B H: B G=5: 14, \therefore \mathrm{S}_{\triangle \mathrm{BEH}}=\frac{5}{14} \times \frac{1}{2} \times \mathrm{S}_{\triangle \mathrm{BFG}}=\frac{1}{42}$,

Therefore, the answer is: $\frac{23}{210}$.