15.8 (League 2001) $D, E$ are points on side $BC$ of $\triangle ABC$, $F$ is a point on the extension of $BA$, $\angle DAE = \angle CAF$.
(1) Determine the positional relationship between the circumcircle of $\triangle ABD$ and the circumcircle of $\triangle AEC$, and prove your conclusion.
(2) If the circumradius of $\triangle ABD$ is twice that of $\triangle AEC$, $BC=6, AB=4$, find the length of $BE$.

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(1) To determine the positional relationship between the circumcircle of $\triangle ABD$ and the circumcircle of $\triangle AEC$, we need to analyze the given angles and the geometric properties of the circles.

Given:
- $\angle DAE = \angle CAF$
- $D$ and $E$ are on $BC$
- $F$ is on the extension of $BA$

Since $\angle DAE = \angle CAF$, we can infer that the points $A, D, E, F$ are concyclic. This means that the circumcircle of $\triangle ADE$ passes through $F$.

Now, consider the circumcircle of $\triangle ABD$ and the circumcircle of $\triangle AEC$. Since $\angle DAE = \angle CAF$, the points $A, D, E, F$ lie on a common circle. Therefore, the circumcircle of $\triangle ABD$ and the circumcircle of $\triangle AEC$ intersect at points $A$ and $E$.

Thus, the circumcircle of $\triangle ABD$ and the circumcircle of $\triangle AEC$ are **intersecting** at points $A$ and $E$.

(2) Given that the circumradius of $\triangle ABD$ is twice that of $\triangle AEC$, and $BC = 6$, $AB = 4$, we need to find the length of $BE$.

Let the circumradius of $\triangle AEC$ be $R$ and the circumradius of $\triangle ABD$ be $2R$.

Using the circumradius formula for a triangle:
\[ R = \frac{abc}{4K} \]
where $a, b, c$ are the sides of the triangle and $K$ is the area of the triangle.

For $\triangle AEC$:
\[ R = \frac{AE \cdot AC \cdot EC}{4K_{AEC}} \]

For $\triangle ABD$:
\[ 2R = \frac{AB \cdot AD \cdot BD}{4K_{ABD}} \]

Since $AB = 4$ and $BC = 6$, let $BE = x$ and $EC = 6 - x$.

Using the fact that the circumradius of $\triangle ABD$ is twice that of $\triangle AEC$, we can set up the following relationship:
\[ 2 \left( \frac{AE \cdot AC \cdot EC}{4K_{AEC}} \right) = \frac{4 \cdot AD \cdot BD}{4K_{ABD}} \]

Simplifying, we get:
\[ 2 \left( \frac{AE \cdot AC \cdot (6 - x)}{4K_{AEC}} \right) = \frac{4 \cdot AD \cdot (x + 4)}{4K_{ABD}} \]

Since the areas $K_{AEC}$ and $K_{ABD}$ are proportional to the product of the sides, we can simplify further:
\[ 2 \left( \frac{AE \cdot AC \cdot (6 - x)}{4K_{AEC}} \right) = \frac{4 \cdot AD \cdot (x + 4)}{4K_{ABD}} \]

Given that the circumradius relationship holds, we can solve for $x$:
\[ 2 \left( \frac{AE \cdot AC \cdot (6 - x)}{4K_{AEC}} \right) = \frac{4 \cdot AD \cdot (x + 4)}{4K_{ABD}} \]

Since the areas are proportional, we can simplify to:
\[ 2(6 - x) = x + 4 \]

Solving for $x$:
\[ 12 - 2x = x + 4 \]
\[ 12 - 4 = 3x \]
\[ 8 = 3x \]
\[ x = \frac{8}{3} \]

Thus, the length of $BE$ is:
\[ \boxed{\frac{8}{3}} \]
15. 8 (1) Draw the tangent $AH$ of $\odot ABD$ such that $\angle DAE = \angle DAH + \angle HAE$, then $\angle CAF = \angle B + \angle C = \angle DAH + \angle C, \therefore \angle HAE = \angle C, AH$ is also the tangent of $\odot AEC$. Therefore, the two circles are tangent at point $A$.
(2) Extend $DA$ to intersect $\odot AEC$ at $G$, and extend $BA$ to intersect $\odot AEC$ at $F$. It can be proven that $\angle ADB = \angle FAC + \angle AED = \angle FAC + \angle AFC = \angle AGF$, hence $\triangle ADB \sim \triangle AGF$. Given $AB:AF = 2$, thus $AF = 2$. Using the intersecting chords theorem, we find $BE = 4$.