In parallelogram $A B C D$, $E, F$ are on $A D, A B$ respectively. It is known that $S_{A F I E}=49, S_{\triangle B G F}=13, S_{\triangle D E H}=35$, find $S_{G C H I}=$ $\qquad$.


【Analysis and Solution】
Geometry, half model.
(Method 1)
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\begin{array}{l}
S_{\text {quadrilateral } G C H I}+\left(S_{\triangle E H I}+S_{\triangle B C G}\right)=S_{\triangle B C E}=S_{\text {parallelogram } A B C D} \div 2 ; \\
\left(S_{\text {quadrilateral } A F I E}+S_{\triangle B G F}+S_{\triangle D E H}\right)+\left(S_{\triangle E H I}+S_{\triangle B C G}\right)=S_{\triangle A D F}+S_{\triangle B C F}=S_{\text {parallelogram } A B C D} \div 2 ; \\
S_{\text {quadrilateral } G C H I}+\left(S_{\triangle E H I}+S_{\triangle B C G}\right)=\left(S_{\text {quadrilateral } A F I E}+S_{\triangle B G F}+S_{\triangle D E H}\right)+\left(S_{\triangle E H I}+S_{\triangle B C G}\right) ; \\
S_{\text {quadrilateral } G C H I}=S_{\text {quadrilateral } A F I E}+S_{\triangle B G F}+S_{\triangle D E H}=49+13+35=97 . \\
(\text { Method 2 }) \\
S_{\text {quadrilateral } G C H I}+\left(S_{\triangle F G I}+S_{\triangle C D H}\right)=S_{\triangle C D F}=S_{\text {parallelogram } A B C D} \div 2 ; \\
\left(S_{\text {quadrilateral } A F I E}+S_{\triangle B G F}+S_{\triangle D E H}\right)+\left(S_{\triangle F G I}+S_{\triangle C D H}\right)=S_{\triangle A B E}+S_{\triangle C D E}=S_{\text {parallelogram } A B C D} \div 2 ; \\
S_{\text {quadrilateral } G C H I}+\left(S_{\triangle F G I}+S_{\triangle C D H}\right)=\left(S_{\text {quadrilateral } A F I E}+S_{\triangle B G F}+S_{\triangle D E H}\right)+\left(S_{\triangle F G I}+S_{\triangle C D H}\right) ; \\
S_{\text {quadrilateral } G C H I}=S_{\text {quadrilateral } A F I E}+S_{\triangle B G F}+S_{\triangle D E H}=49+13+35=97 .
\end{array}
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