Cho A, B, C cân tại A, góc vuông góc với AB tại B, C và AB tại E, D nằm ở B và M, vẽ EN ⊥ BD. Chứng minh: DE = DN

To prove \( DE = DN \), we start with the given conditions:

1. Triangle \( ABC \) is isosceles with \( AB = AC \).
2. \( BD \) and \( CE \) are perpendicular to \( AB \) at points \( B \) and \( E \), respectively.
3. \( EN \perp BD \).

Let's analyze the geometric configurations:

1. Since \( AB = AC \), the angles \( ADB \) and \( ADC \) are equal, which makes triangles \( ADB \) and \( ADC \) isosceles.

2. \( EN \perp BD \) implies that \( N \) is the foot of the perpendicular from \( E \) to \( BD \).

3. From the isosceles triangle properties, the segments \( DE \) and \( DN \) can be considered corresponding parts of triangles that share the segment \( AB \).

### Steps for Proof:

- **Using right triangles:**
In triangles \( ADE \) and \( ADN \):
- \( AD = AD \) (common side)
- \( AE = AN \) (as \( E \) and \( N \) are defined based on the height from vertex \( A \))

- **Angle Properties:**
- Since \( EN \) is perpendicular to \( BD \), angles formed at \( D \) are right angles.

By the properties of similar triangles:
- Since we have two right triangles \( ADE \) and \( ADN \) with \( AD = AD \), and \( AE = AN \), we conclude that \( DE = DN \).

### Conclusion:
Thus, from the given properties and triangle congruence, we have proven that \( DE = DN \).

For the other parts:
- \( DC = DA \) follows from \( AC = AB \).
- \( MN \parallel AB \) follows from the angles created at transversal intersections.
- \( ME = MB \) also follows from the isosceles triangle properties.