Problem 1325

1. Identify the given information and the goal:
- We have a triangle $ABC$ with sides $AB = 7$, $BC = 8$, and $AC = 6$.
- $D$ is the midpoint of $BC$, so $BD = DC = \frac{8}{2} = 4$.
- We need to find the length of $AE$ where $E$ is the point where the circle through $A$, $B$, and $D$ intersects $AC$ again.

2. Apply the Power of a Point theorem:
- The Power of a Point theorem states that for a point $C$ outside a circle, if a line through $C$ intersects the circle at points $E$ and $A$, then $CE \cdot CA = CD \cdot CB$.

3. Substitute the known values into the Power of a Point equation:
$$CE \cdot CA = CD \cdot CB$$
$$CE \cdot 6 = 4 \cdot 8$$

4. **Solve for $CE$:**
$$CE \cdot 6 = 32$$
$$CE = \frac{32}{6} = \frac{16}{3}$$

5. **Determine $AE$ using the segment subtraction:**
- Since $E$ is on $AC$, we have $AE = AC - CE$.
$$AE = 6 - \frac{16}{3}$$
$$AE = \frac{18}{3} - \frac{16}{3} = \frac{2}{3}$$

The final answer is $\boxed{\frac{2}{3}}$