【Question 24】
As shown in the figure, point $E$ is any point on the diagonal $AC$ of rectangle $ABCD$. A perpendicular line from $E$ to $AB$ and $BC$ intersects $AB$ and $BC$ at points $F$ and $G$, respectively. Connect $DF$, $FG$, and $GD$. Given that $AB=8$, $AD=10$, and the area of triangle $DFG$ is 30, then the area of rectangle $BGEF$ is $\qquad$ $-$.
【Analysis and Solution】
Geometry, Area.
Extend $FE$, $GE$ to intersect $CD$, $AD$ at points $H$, $I$ respectively; connect $DE$;
Since $S_{\triangle DEF}=S_{\triangle AEF}$, $S_{\triangle DEG}=S_{\triangle CEG}$;
Therefore, $S_{\triangle DEF}+S_{\triangle DEG}+S_{\triangle EFG}=S_{\triangle AEF}+S_{\triangle CEG}+S_{\triangle EFG}$, which means $S_{\triangle DFG}=S_{\text{quadrilateral } AFGC}$;
Since $S_{\triangle DFG}=30$; thus $S_{\text{quadrilateral } AFGC}=30$;
Since $S_{\triangle ABC}=AC \times AB \div 2=10 \times 8 \div 2=40$;
Therefore, $S_{\triangle BFG}=S_{\triangle ABC}-S_{\text{quadrilateral } AFGC}=40-30=10$;
Therefore, $S_{\text{rectangle } BGEF}=S_{\triangle BFG} \times 2=10 \times 2=20$.