Let $ABC$ be an isosceles triangle with $AB = BC$. Point $E$ lies on the side $AB$, and $ED$ is the perpendicular from $E$ to $BC$. It is known that $AE = DE$. Find $\angle DAC$.
1. **Identify the given information and set up the problem:**
   - Triangle \(ABC\) is isosceles with \(AB = BC\).
   - Point \(E\) lies on side \(AB\).
   - \(ED\) is the perpendicular from \(E\) to \(BC\).
   - \(AE = DE\).
   - We need to find \(\angle DAC\).

2. **Introduce angles and use properties of isosceles triangles:**
   - Let \(\angle BAD = \alpha\) and \(\angle DAC = \beta\).
   - Since \(\triangle AED\) is isosceles with \(AE = DE\), we have \(\angle ADE = \angle EAD = \alpha\).

3. **Determine \(\angle CDA\):**
   - Since \(ED\) is perpendicular to \(BC\), \(\angle EDC = 90^\circ\).
   - Therefore, \(\angle CDA = 90^\circ - \alpha\).

4. **Use the property of isosceles triangle \(ABC\):**
   - Since \(AB = BC\), \(\angle ACB = \angle BAC\).
   - Let \(\angle ACB = \gamma\). Then \(\gamma = \alpha + \beta\).

5. **Express \(\angle ACB\) in terms of \(\alpha\) and \(\beta\):**
   - From the previous step, \(\angle ACB = 90^\circ + \alpha - \beta\).

6. **Set up the equation using the equality of angles in isosceles triangle:**
   - Since \(\angle ACB = \angle BAC\), we have:
     \[
     90^\circ + \alpha - \beta = \alpha + \beta
     \]

7. **Solve for \(\beta\):**
   \[
   90^\circ + \alpha - \beta = \alpha + \beta
   \]
   \[
   90^\circ = 2\beta
   \]
   \[
   \beta = 45^\circ
   \]

8. **Conclude the value of \(\angle DAC\):**
   \[
   \angle DAC = \beta = 45^\circ
   \]

The final answer is \(\boxed{45^\circ}\)