Example 5 As shown in Figure 9, given a square $A B C D$ divided into four smaller rectangles by two line segments $E F$ and $G H$ parallel to the sides, $P$ is the intersection of $E F$ and $G H$.
If the area of rectangle $P F C H$ is exactly twice the area of rectangle $A G P E$, determine the size of $\angle H A F$ and prove your conclusion.
(1998, Beijing Middle School Mathematics Competition (Junior

2))
Solution: Guess $\angle F A H=45^{\circ}$. Proof as follows:
Let $A G=a, B G=b, A E=x, E D=y$. Then $a+b=x+y, 2 a x=b y$.
Thus, $a-x=y-b$, which means $a^{2}-2 a x+x^{2}=y^{2}-2 b y+b^{2}=y^{2}-4 a x+b^{2}$. Therefore, $(a+x)^{2}=b^{2}+y^{2}$, which means $a+x=\sqrt{b^{2}+y^{2}}=F H$.
Hence, $D H+B F=F H$.
Extend $C B$ to point $M$, such that $B M=D H$, and connect $A M$. By Rt $\triangle A B M \cong \mathrm{Rt} \triangle A D H$, we get $A M=A H, \angle M A B=\angle H A D$. Therefore, $\angle M A H=\angle M A B+\angle B A H$ $=\angle B A H+\angle H A D=90^{\circ}$.
Furthermore, by $\triangle A M F \cong \triangle A H F$, we know $\angle M A F=\angle H A F$.
Thus, $\angle H A F=\frac{1}{2} \angle M A H=45^{\circ}$.