Cho tam giác ABC đường cao BD; CE cắt nhau ở H, AH cắt BC tại M

The problem you provided involves a triangle \( ABC \) with height \( BD \) and intersection points involving various lines. Here's a breakdown of the components and tasks:

1. **Given:**
- Triangle \( ABC \) where \( BD \) is the altitude.
- Line \( CE \) intersects \( BD \) at point \( H \) and line \( AH \) intersects \( BC \) at point \( M \).

2. **To prove:**
- \( AM \perp BC \)
- \( \angle BAM = \angle ECB \)

3. **Approach:**
- Use properties of triangle and angles to prove the statements.
- For part (b), establish \( k \) such that triangle \( AB \) is similar to triangle \( HK \), which may involve angle bisector theorem or similarity arguments.

Would you like a more detailed explanation or solution for one of the sections?