Cho Δ ABC vuông tại A (AC > AB), đường cao AH

To solve the problem, we start with triangle \( \Delta ABC \) where \( \angle A \) is the right angle, and we have \( AC > AB \). The height from \( A \) to \( BC \) is denoted as \( AH \).

1. **Given Information**:
- \( AM = MI \)
- Right triangle \( \Delta AHI \) with \( AI \) being the height.

**Questions**:
a) Prove that \( \Delta ACI \sim \Delta ABN \) (similar triangles)
b) Prove that \( ANT = ACI \) (congruent triangles)
c) Prove that \( MI \perp NI \) (right angles)

### Solutions:

a) **Similarity \( \Delta ACI \sim \Delta ABN \)**:
- Since \( \angle A \) is common and both triangles share \( AH \) as a height (perpendicular), we can use the Angle-Angle (AA) criterion for similarity.

b) **Congruence \( ANT = ACI \)**:
- Since triangles have equal angles and one side \( AN \) that corresponds to side \( AC \) and is equal, we can conclude by the Side-Angle-Side (SAS) postulate.

c) **Right angles \( MI \perp NI \)**:
- Since \( MI \) and \( NI \) are both heights from points to the base \( BC \), they are perpendicular to \( BC \).

### Conclusion:
- Using the properties of right triangles and similarity/congruence criteria, we can justify each statement based on the properties of triangles in this scenario.