61. As shown in the figure, three equilateral triangles: $\triangle A B C, \triangle H F G, \triangle D C E$ are placed on the line $m$. It is known that $B C=\frac{1}{3} C E, F, G$ are the midpoints of $B C, C E$ respectively, $F M / / A C, G N / / D C$. Let the areas of the three parallelograms in the figure be $S_{1}, S_{2}, S_{3}$ respectively. If $S_{1}+S_{3}=10$, then $S_{2}=$ . $\qquad$
Answer: 3
Solution: According to the properties of an equilateral triangle, we have
$$
\angle A B C=\angle H F G=\angle D C E=60^{\circ} \text {, }
$$

Therefore,
$$
A B / / H F / / D C / / G N,
$$

Let $A C$ intersect $F H$ at $P$, and $C D$ intersect $H G$ at $Q$,
then $\triangle M B F, \triangle P F C, \triangle Q C G$ and $\triangle N G E$ are equilateral triangles,
since $F, G$ are the midpoints of $B C, C E$ respectively, and $B C=\frac{1}{3} C E$,
thus
$$
F G=\frac{1}{2} B C+\frac{1}{2} C E=2 B C \text {. }
$$

By the area formula of a triangle, we know
$$
S_{\triangle C D E}=9 S_{\triangle A B C}, S_{\triangle F H G}=4 S_{\triangle A B C} \text {, }
$$

Since $F, G$ are the midpoints of $B C, C E$ respectively, we have

Therefore,
$$
\begin{array}{c}
S_{\triangle B M F}=S_{\triangle F P C}=\frac{1}{4} S_{\triangle A B C} \\
S_{\triangle C Q G}=S_{\triangle G N E}=\frac{1}{4} S_{\triangle C D E} \\
S_{1}=\left(1-\frac{1}{4} \times 2\right) S_{\triangle A B C}=\frac{1}{2} S_{\triangle A B C} \\
S_{3}=\left(1-\frac{1}{4} \times 2\right) S_{\triangle C D E}=\frac{9}{2} S_{\triangle A B C} \\
S_{1}+S_{3}=\frac{1}{2} S_{\triangle A B C}+\frac{9}{2} S_{\triangle A B C}=5 S_{\triangle A B C}=10
\end{array}
$$

Therefore,
$$
\begin{array}{l} 
S_{\triangle A B C}=2 \\
S_{2}= S_{\triangle F H G}-\frac{1}{4} S_{\triangle A B C}-\frac{1}{4} S_{\triangle C D E} \\
= 4 S_{\triangle A B C}-\frac{1}{4} S_{\triangle A B C}-\frac{9}{4} S_{\triangle A B C} \\
= \frac{3}{2} S_{\triangle A B C} \\
= 3 .
\end{array}
$$