Example 1 As shown in Figure 1, cut off three corners (remove three isosceles right triangles) from a square paper with a side length of 1. How can the remaining part be divided into three parts along two straight lines, and then the three parts be joined to form a square.
First, consider the side length of the new square formed? The side length of the original square is 1, so the area of the remaining pentagon $A C D E F$, as shown in Figure 2, is
$$
1-3 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{5}{8} \text {. }
$$

Thus, the side length of the new square is $\sqrt{\frac{5}{8}}$. Next, we try to determine a point $B$ on the side $C D$ such that $A B$ is the side length of the new square.

Since
$$
A C=\frac{1}{\sqrt{2}}<\sqrt{\frac{5}{8}}<1=A D .
$$

And $C B=\sqrt{\frac{5}{8}-\frac{1}{2}}=\frac{1}{2} C D$.
Therefore, $B$ is the midpoint of $C D$.
Cut along the line $A B$. Remove the right triangle $\triangle A C B$, and place it so that $A C$ coincides with $A F$, i.e., in the position of $\triangle A F B^{\prime}$ in Figure 2. $\angle E F B^{\prime}$ $=135^{\circ}$, which is exactly the degree of $\angle B D E$. The second cut is along $B E$, and move $\triangle B D E$ to the position of $\triangle B^{\prime} F E$, forming a perfect square $A B E B^{\prime}$. Proof is omitted.