Cho ΔABC vuông tại A (AB < AC), đường cao AH, đường trung tuyến AM

To solve the problem related to triangle \( \triangle ABC \) right-angled at \( A \), follow these steps:

1. **Given Values**:
- Length of \( BC = 10 \, \text{cm} \)
- Height \( BH = 3.6 \, \text{cm} \)

2. **Calculate Other Lengths**:
- **Using Pythagorean Theorem**: Since \( \triangle ABC \) is right-angled at \( A \), we can find the lengths of \( AB \) and \( AC \).
- Let \( AB = a \) and \( AC = b \). From \( BH = 3.6 \, \text{cm} \), we can deduce that:
\[
AH = \frac{a \cdot b}{c} \quad \text{(where \( c \) is the hypotenuse \( BC = 10 \, \text{cm} \))}
\]

3. **Finding \( AH \)**:
\[
AH = \frac{AB \cdot AC \cdot \sin C}{BC} = \frac{AB \cdot AC \cdot \sin(90^\circ)}{10} = \frac{AB \cdot AC}{10}
\]

4. **Determine Ratios**:
- Given the relationship in triangles and using the area approach, you can compare the areas to set up ratios:
- The area of \( \triangle ABH \) and \( \triangle AHC \) can give insights into other segment lengths.

5. **Using Perpendiculars and Similarities**:
- Establish the triangle \( BE \) which is perpendicular to \( AM \). Note that because \( D \) lies on \( AH \), this suggests the properties of triangle similarity you could use.

6. **Positioning Point K**:
- For point \( K \) on segment \( BE \) such that \( AKM = 90^\circ \), you can analyze \( \triangle AKM \):
- Show that the areas relate as described and use trigonometric identities if \( AK \) and \( MK \) interact as proposed.

7. **Final Relations**:
- Set up the equations to show \( AE = BE \) and confirm using \( S_{AAMB} \) and \( S_{AMCD} \) to finalize area relationships.

By following these steps, you'll derive all required lengths, ratios, and final equational relationships necessary for \( AB \) and \( AC \) in \( \triangle ABC \).