In triangle $ABC$, $AB = 28$, $AC = 36$, and $BC = 32$. Let $D$ be the point on segment $BC$ satisfying $\angle BAD = \angle DAC$, and let $E$ be the unique point such that $DE \parallel AB$ and line $AE$ is tangent to the circumcircle of $ABC$. Find the length of segment $AE$.

[i]Ray Li[/i]
1. **Identify the given lengths and properties:**
   - In triangle \(ABC\), we have \(AB = 28\), \(AC = 36\), and \(BC = 32\).
   - Point \(D\) on segment \(BC\) satisfies \(\angle BAD = \angle DAC\), making \(AD\) the angle bisector of \(\angle BAC\).

2. **Use the Angle Bisector Theorem:**
   - The Angle Bisector Theorem states that the angle bisector of an angle in a triangle divides the opposite side into segments proportional to the adjacent sides. Therefore,
   \[
   \frac{BD}{DC} = \frac{AB}{AC} = \frac{28}{36} = \frac{7}{9}
   \]
   - Let \(BD = x\) and \(DC = 32 - x\). Then,
   \[
   \frac{x}{32 - x} = \frac{7}{9}
   \]
   - Solving for \(x\):
   \[
   9x = 7(32 - x) \implies 9x = 224 - 7x \implies 16x = 224 \implies x = 14
   \]
   - Thus, \(BD = 14\) and \(DC = 32 - 14 = 18\).

3. **Establish the parallel line and cyclic quadrilateral:**
   - Since \(DE \parallel AB\), \(\angle CDE = \angle CBA\).
   - Given that \(AE\) is tangent to the circumcircle of \(\triangle ABC\), \(\angle BAE = \angle ACB\) (tangent-secant angle theorem).

4. **Prove cyclic quadrilateral and congruence:**
   - Since \(\angle BAD = \angle DAC\), \(\angle CAD = \angle EAD\).
   - \(\angle CDE = \angle CBA\) implies that quadrilateral \(AECD\) is cyclic.
   - In cyclic quadrilateral \(AECD\), \(\angle CAD = \angle EAD\) and \(\angle CDE = \angle CBA\).
   - By ASA (Angle-Side-Angle) Congruence Postulate, \(\triangle CAD \cong \triangle EAD\).

5. **Determine the length of \(AE\):**
   - Since \(\triangle CAD \cong \triangle EAD\), corresponding sides are equal, so \(AE = CD\).
   - We have already found \(CD = 18\).

The final answer is \(\boxed{18}\).