9. (30 points) On the side $B C$ of the square $A B C D$, points $E$ and $F$ are marked such that $B E: E C = C F: F B = 1: 2$. On the side $C D$, point $G$ is marked such that $C G: G D = 2: 1$. On the side $A D$, points $H$ and $I$ are marked such that $A I: I D = D H: H A = 1: 2$. Segment $B G$ intersects segments $A E, I F$, and $H C$ at points $J, K$, and $L$ respectively. Which of the quadrilaterals has a larger area - EFKJ or GDHL?
Answer: The area of $G D H L$ is larger.

![](https://cdn.mathpix.com/cropped/2024_05_06_f23fd9be54420f8c770fg-5.jpg?height=462&width=619&top_left_y=1134&top_left_x=707)

Solution: Let $S_{A B C D}=36 S, S_{C L G}=x$. Then $S_{C D H}=6 S$ and $S_{G D H L}=6 S-x$, $S_{B C G}=12 S$ and $S_{B C L}=12 S-x$. From the equalities $B E=E F=F C=A I=I H=H D$ it follows that $A E\|I F\| C H$, and by Thales' theorem $B J=J K=K L$. Therefore, triangles $B J E$, $B K F$, and $B L C$ are similar. Next, we find $S_{B J E}=\frac{1}{9} S_{B C L}, S_{B K F}=\frac{4}{9} S_{B C L}$ and $S_{E F K J}=S_{B K F}-$ $-S_{B J E}=\frac{1}{3} S_{B C L}=4 S-\frac{x}{3}$.

To compare the areas of the given quadrilaterals, it remains to find $x$. Let $N$ be the midpoint of side $B C$. Extend side $B C$ beyond point $B$ and mark a point $M$ such that $C N=N B=B M$. On side $C D$, mark point $O$ - the midpoint of segment $C G$, that is, $C O=O G=G D$. Then $N O\|B G\| M D$, and by Thales' theorem $C P=$ $=P L=L Q$. Consider triangles $Q H D$ and $Q M C$. They are similar with a coefficient equal to $\frac{H D}{C M}=\frac{2}{9}=\frac{H Q}{Q C}$. Consequently, $\frac{C L}{L H}=\frac{6}{5}$. Then $x=\frac{6}{11} \cdot \frac{2}{3} S_{C H D}=\frac{4}{11} \cdot 6 S=\frac{24}{11} S$, from which $S_{G D H L}=\frac{42}{11} S, S_{E F K J}=\frac{36}{11} S$.