8. (10 points) As shown in the figure, a square piece of paper $A B C D$ is cut along the diagonal $B D$ into two triangles. In the first operation, triangle $A B D$ is moved vertically downward by 3 cm to triangle $E F G$; in the second operation, triangle $E F G$ is moved vertically downward by another 5 cm to triangle $H I J$. The overlapping area of the two pieces of paper after the first operation is equal to the overlapping area of the two pieces of paper after the second operation. What is the area of the square piece of paper $A B C D$ in square centimeters? $\qquad$
【Analysis】The first overlapping part is parallelogram $K B N G$, and the second overlapping part is parallelogram $B O J L$. These two parts have equal areas. By subtracting parallelogram $B N M L$ from both, we get that parallelogram $K L M G$ and parallelogram $M N O J$ have equal areas.

【Solution】
Parallelogram $K L M G = 5 \times 3 = 15$ (square centimeters)
Since all the triangles in the figure are isosceles right triangles, $B I = B O = 3 + 5$, $B F = B N = 3$, so $N O = 5$ centimeters
$J C = 15 \div 5 = 3$ (centimeters)
The side length of the square is $3 + 5 + 3 = 11$ (centimeters)
The area of the square is $11 \times 11 = 121$ (square centimeters)
Therefore, the answer is 121.