In $\Delta ABC$, $AC=kAB$, with $k>1$. The internal angle bisector of $\angle BAC$ meets $BC$ at $D$. The circle with $AC$ as diameter cuts the extension of $AD$ at $E$. Express $\dfrac{AD}{AE}$ in terms of $k$.
1. **Identify the given information and setup the problem:**
   - In $\Delta ABC$, $AC = kAB$ with $k > 1$.
   - The internal angle bisector of $\angle BAC$ meets $BC$ at $D$.
   - The circle with $AC$ as diameter cuts the extension of $AD$ at $E$.
   - We need to express $\dfrac{AD}{AE}$ in terms of $k$.

2. **Construct the point $F$ on $AD$ such that $BF \perp AD$:**
   - This construction helps us use similar triangles to find the necessary ratios.

3. **Identify similar triangles:**
   - $\triangle ABF \sim \triangle ACE$ because $\angle ABF = \angle ACE = 90^\circ$ and $\angle BAF = \angle CAE$.
   - $\triangle DBF \sim \triangle DCE$ because $\angle DBF = \angle DCE = 90^\circ$ and $\angle BDF = \angle CDE$.

4. **Write the ratios from the similar triangles:**
   - From $\triangle ABF \sim \triangle ACE$, we have:
     \[
     \dfrac{AF}{AE} = \dfrac{AB}{AC} = \dfrac{1}{k}
     \]
   - From $\triangle DBF \sim \triangle DCE$, we have:
     \[
     \dfrac{DF}{DE} = \dfrac{DB}{DC} = \dfrac{1}{k}
     \]

5. **Express lengths in terms of $x$ and $y$:**
   - Let $AF = x$ and $AE = xk$.
   - Let $DF = y$ and $DE = yk$.

6. **Use the segment addition to find a relationship:**
   - Since $AF + FD + DE = AE$, we have:
     \[
     x + y + yk = xk
     \]
   - Rearrange to solve for $\dfrac{y}{x}$:
     \[
     x + y + yk = xk \implies 1 + \dfrac{y}{x} + \dfrac{y}{x}k = k \implies \dfrac{y}{x}(1 + k) = k - 1 \implies \dfrac{y}{x} = \dfrac{k - 1}{k + 1}
     \]

7. **Find the desired ratio $\dfrac{AD}{AE}$:**
   - Since $AD = AF + FD = x + y$, we have:
     \[
     \dfrac{AD}{AE} = \dfrac{x + y}{xk} = \dfrac{x}{xk} + \dfrac{y}{xk} = \dfrac{1}{k} + \dfrac{y}{xk}
     \]
   - Substitute $\dfrac{y}{x} = \dfrac{k - 1}{k + 1}$:
     \[
     \dfrac{AD}{AE} = \dfrac{1}{k} + \dfrac{k - 1}{k(k + 1)} = \dfrac{1}{k} + \dfrac{k - 1}{k^2 + k} = \dfrac{1}{k} + \dfrac{k - 1}{k^2 + k}
     \]
   - Combine the fractions:
     \[
     \dfrac{AD}{AE} = \dfrac{k + k - 1}{k(k + 1)} = \dfrac{2k - 1}{k(k + 1)} = \dfrac{2}{k + 1}
     \]

The final answer is $\boxed{\dfrac{2}{k + 1}}$