7.(40 points) As shown in Figure 7, quadrilateral $ABCD$ is inscribed in a circle, $AB=AD$, and its diagonals intersect at point $E$. Point $F$ lies on segment $AC$ such that $\angle BFC = \angle BAD$. If $\angle BAD = 2 \angle DFC$, find the value of $\frac{BE}{DE}$.

保留源文本的换行和格式，直接输出翻译结果。
7. From $AB=AD$, we know $\angle ABD=\angle ADB=\theta$. By the property of equal arcs subtending equal angles at the circumference, we have
$$
\angle ACD=\angle ACB=\theta.
$$

Let $\angle DFC=\varphi$. Then
$$
\angle BAD=\angle BFC=2\varphi.
$$

Therefore, $\angle ABD + \angle ADB + \angle BAD$
$$
=\theta + \theta + 2\varphi = 180^{\circ}.
$$

Thus, $\theta + \varphi = 90^{\circ}$, and $\angle CDF = 90^{\circ}$.
On the other hand, from
$$
\begin{array}{l}
\angle FBC = 180^{\circ} - \theta - 2\varphi = \theta = \angle FCB \\
\Rightarrow FB = FC.
\end{array}
$$

Let $M$ be the midpoint of side $BC$, and connect $FM$. It is easy to see that $\triangle FCD \cong \triangle FBM$, and $BC = 2CD$.
Since $AC$ is the angle bisector of $\angle BCD$, by the Angle Bisector Theorem, we have $\frac{BE}{DE} = \frac{BC}{CD} = 2$.