Let $ABC$ be a triangle such that $|AB|=7$, $|BC|=8$, $|AC|=6$. Let $D$ be the midpoint of side $[BC]$. If the circle through $A$, $B$ and $D$ cuts $AC$ at $A$ and $E$, what is $|AE|$?

$ 
\textbf{(A)}\ \dfrac 23
\qquad\textbf{(B)}\ 1
\qquad\textbf{(C)}\ \dfrac 32
\qquad\textbf{(D)}\ 2
\qquad\textbf{(E)}\ 3
$
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 \(\boxed{\frac{2}{3}}\)