Kẻ HM vuông góc AB tại M và HM vuông góc AC tại N. Chứng minh tan^3C = BM/CN

To prove \( \tan^3 C = \frac{BM}{CN} \) based on the given information in the triangle \( \triangle ABC \):

1. **Understanding the Configuration**:
- Given a right triangle \( \triangle ABC \) at \( A \), with \( AH \) being the height from \( A \) to the base \( BC \).
- \( AE = 4 \, \text{cm} \) and \( \angle C = 40^\circ \).
- We drop perpendiculars \( HM \) from \( H \) to \( AB \) at point \( M \) and \( HN \) from \( H \) to \( AC \) at point \( N \).

2. **Using Trigonometric Relationships**:
- From the right triangle \( \triangle AHC \):
\[
AH = AC \cdot \sin C = 4 \cdot \sin 40^\circ
\]

3. **Finding Lengths**:
- For \( BM \) and \( CN \):
- Use the sine and cosine relationships to express these lengths in terms of \( \tan C \):
\[
BM = AB \cdot \cos C \quad \text{and} \quad CN = AC \cdot \sin C
\]

4. **Calculate \( BM \) and \( CN \)**:
- Substitute \( \tan C \):
\[
\tan C = \frac{AH}{BM} = \frac{4 \cdot \sin 40^\circ}{AB \cdot \cos 40^\circ}
\]
- This would give \( BM = \frac{4 \cdot \sin 40^\circ \cdot \cos 40^\circ}{\sin C} \).

5. **Then, using the ratio**:
- Rearrange to isolate \( \frac{BM}{CN} \):
\[
\frac{BM}{CN} = \frac{AB \cdot \cos(40^\circ)}{AC \cdot \sin(40^\circ)}
\]

6. **Final Expression**:
- Show that \( \tan^3 C \) simplifies to this ratio through the relationships inherent in the triangle.

Through these steps and the relationships outlined, you conclude that \( \tan^3 C = \frac{BM}{CN} \) holds true with geometric reasoning and trigonometric relationships established within the right triangle.