3. In square $A B C D$, points $E$ and $F$ are on sides $A D$ and $B C$ respectively, such that $E F / / A B$; points $G$ and $H$ are on sides $A B$ and $D C$ respectively, such that $G H / / A D$; $E F$ intersects $G H$ at point $K$. If the area of rectangle $K F C H$ is twice the area of rectangle $A G K E$, find the degree measure of $\angle F A H$.


3. As shown in Figure 7, rotate $\triangle A B F$ counterclockwise around point $A$ by $90^{\circ}$, until point $B$ coincides with $D$ and point $F$ coincides with point $P$ on the extension of $H D$.
Let $S_{\text {trapezoid } A B C D}=1$.
Let $A E=x, A G=y$.

From $S_{\text {rectangle } K F C H}=2 S_{\text {trapezoid } A G K E}$
$$
\Rightarrow(1-x)(1-y)=2 x y \text {. }
$$

Then $S_{\triangle A F H}=S_{\triangle A K H}+S_{\triangle A K F}+S_{\triangle K F H}$
$$
\begin{array}{l}
=\frac{S_{\text {rectangle } A B C D}-S_{\text {trapezoid } G K E}}{2} \\
=\frac{1-x y}{2}=\frac{x+y}{2},
\end{array}
$$
$S_{\triangle A H P}=S_{\triangle A B F}+S_{\triangle A D H}=\frac{x+y}{2}$.
Also, $A F=A P, A H=A H$, thus,
$$
\angle F A H=\angle P A H \text {. }
$$

At this moment, $A P$ is the image of $A F$ after a counterclockwise rotation of $90^{\circ}$.
Hence $\angle F A P=90^{\circ} \Rightarrow \angle F A H=45^{\circ}$.