ΔABC ⊥ tại A BM là đường phân giác của ∠ABC (M ∈ AC), M là điểm IAC IM cắt B tại N

To solve the problem involving triangle \( \Delta ABC \) with given conditions, we can follow these steps:

### Given
1. \( \Delta ABC \perp \) at \( A \).
2. \( BM \) is the angle bisector of \( \angle ABC \) (with \( M \in AC \)).
3. \( I \) is the incenter of \( \Delta ABC \).
4. \( IM \) intersects \( B \) at \( N \).

### Required Prove
We need to show the following conclusions:
a. \( \Delta BAI \cong \Delta BMA \)
b. \( \Delta BAI \) is isosceles.
c. \( \Delta AMN \cong \Delta AMC \)
d. \( AI \parallel NC \)
e. \( \Delta BNC \) is isosceles.

### Steps

1. **Construct the Triangle:**
- Start with triangle \( \Delta ABC \) with \( A \) at a right angle.

2. **Properties of the Angle Bisector:**
- Since \( BM \) is the angle bisector, by the Angle Bisector Theorem:
\[ \frac{AM}{MC} = \frac{AB}{BC} \]

3. **Prove Congruence and Similarity:**
- Using properties of the angles at \( A \) being right angles and congruency of corresponding angles, establish:
\[ \Delta BAI \cong \Delta BMA \quad \text{(by AAS - Angle-Angle-Side)} \]
- Given \( AI = AM \) (as both are radii of the incircle).

4. **Isosceles Property:**
- Since \( \Delta BAI \cong \Delta BMA \), we can state that:
\[ \Delta BAI \text{ is isosceles with } BA = BI \]

5. **Further Prove:**
- Using properties of the incenter and intersecting lines, show:
\[ \Delta AMN \cong \Delta AMC \quad \text{(using Angle-Angle postulate)} \]

6. **Parallel Lines:**
- Since \( AI \) bisects \( \angle BAC \) and is parallel to line segment \( NC \), it follows:
\[ AI \parallel NC \]

7. **Final Conclusion:**
- Lastly, show that \( \Delta BNC \) is isosceles using the congruency established in the earlier sections.

> Thus, if all steps hold, we conclude the statements about triangle \( \Delta ABC \) are valid.