In trapezoid $ABCD$, the sides $AD$ and $BC$ are parallel, and $AB = BC = BD$. The height $BK$ intersects the diagonal $AC$ at $M$. Find $\angle CDM$.
1. **Given Information and Initial Setup:**
   - Trapezoid \(ABCD\) with \(AD \parallel BC\).
   - \(AB = BC = BD\).
   - Height \(BK\) intersects diagonal \(AC\) at \(M\).

2. **Objective:**
   - Find \(\angle CDM\).

3. **Prove that Quadrilateral \(BMDC\) is Cyclic:**
   - Let \(\angle CAD = \alpha\).
   - Since \(AB = BC = BD\), triangle \(ABD\) is isosceles with \(AB = BD\).
   - \(\angle DBC = 2\alpha\) because \(B\) is the circumcenter of \(\triangle ADC\).

4. **Isosceles Triangle \(AMD\):**
   - Since \(AD \parallel BC\) and \(AB = BC\), triangle \(AMD\) is isosceles with \(AM = MD\).
   - Therefore, \(\angle DMC = 2\alpha\).

5. **Cyclic Quadrilateral Condition:**
   - For quadrilateral \(BMDC\) to be cyclic, the opposite angles must sum to \(180^\circ\).
   - We have \(\angle DMC = 2\alpha\) and \(\angle DBC = 2\alpha\).
   - Since \(\angle DMC = \angle DBC\), quadrilateral \(BMDC\) is cyclic.

6. **Conclusion:**
   - In a cyclic quadrilateral, the opposite angles sum to \(180^\circ\).
   - Therefore, \(\angle CDM = 90^\circ\).

\(\blacksquare\)

The final answer is \( \boxed{ \angle CDM = 90^\circ } \).