2. The vertex $C$ of squares $A B C D$ and $C J K L$ is an internal point of segments $A K$ and $D J$, points $E$, $F$, $G$, and $H$ are the midpoints of segments $B C$, $B K$, $D K$, and $D C$ respectively. Determine the area of quadrilateral $E F G H$ using the areas $S$ and $T$ of squares $A B C D$ and $C J K L$.
2. Let $a=\sqrt{S}, b=\sqrt{T}$ be the sides of squares $A B C D, C J K L$. Segment $E H$ is the midline of triangle $B C D$ (Fig.1), segment $F G$ is the midline in triangle $B K D$,

![](https://cdn.mathpix.com/cropped/2024_04_17_ba17fecc41b82e1be23bg-2.jpg?height=589&width=577&top_left_y=1176&top_left_x=745)

Fig. 1

thus $2|E H|=2|F G|=|B D|$ and segments $E H, F G$ are parallel to $B D$. Similarly, segment $H G$ is the midline in triangle $D C K$ and segment $E F$ is the midline in triangle $B C K$. Therefore, $2|H G|=2|E F|=|C K|$ and segments $H G, E F$ are parallel to $C K$, and thus perpendicular to $J L$ and $B D$. The parallelogram $E F G H$ is therefore a rectangle with area

$$
|E F| \cdot|F G|=a \frac{1}{2} \sqrt{2} \cdot b \frac{1}{2} \sqrt{2}=\frac{1}{2} a b=\frac{1}{2} \sqrt{S T}
$$

For a complete solution, award 6 points. For discovering that quadrilateral $E F G H$ is a rectangle, award 3 points, and another 3 points for determining its area.