In the quadrilateral $ABCD$, $AB = BC = CD$ and $\angle BMC = 90^\circ$, where $M$ is the midpoint of $AD$. Determine the acute angle between the lines $AC$ and $BD$.
1. **Identify the given information and setup the problem:**
   - Quadrilateral \(ABCD\) with \(AB = BC = CD\).
   - \(\angle BMC = 90^\circ\), where \(M\) is the midpoint of \(AD\).

2. **Define midpoints and properties:**
   - Let \(N\) be the midpoint of \(BC\).
   - Let \(U\) be the midpoint of \(AC\).
   - Let \(V\) be the midpoint of \(BD\).

3. **Establish the properties of the parallelogram \(MUNV\):**
   - Since \(NU = NV = \frac{1}{2} AB\), the parallelogram \(MUNV\) is a rhombus.
   - Given \(\angle BMC = 90^\circ\), we have \(NM = \frac{1}{2} BC = NV = NU\).

4. **Analyze the triangles \(MUN\) and \(MVN\):**
   - Since \(NM = NU\) and \(NM = NV\), triangles \(MUN\) and \(MVN\) are equilateral.
   - Therefore, \(\angle UNV = 120^\circ\).

5. **Determine the angle \(\angle BQC\):**
   - Let \(Q\) be the intersection of \(AB\) and \(DC\).
   - Since \(\angle UNV = 120^\circ\), we have \(\angle BQC = 120^\circ\).

6. **Find the acute angle between \(AC\) and \(BD\):**
   - Let \(P\) be the intersection of the diagonals \(AC\) and \(BD\).
   - Using angle chasing, we find \(\angle CPD\):
     \[
     \angle CPD = \angle CBD + \angle BCA = \frac{1}{2} (\angle BCQ + \angle CBQ) = 30^\circ.
     \]

The final answer is \(\boxed{30^\circ}\).