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

To solve the problem step-by-step, let's break it down based on the information provided:

1. **Given:**
- Triangle \(ABC\) is a right triangle at \(A\) with \(AB < AC\).
- \(BC = 10 \, \text{cm}\), \(BH = 3.6 \, \text{cm}\).
- \(AH\) is the height from \(A\) to \(BC\) and \(AM\) is the median to side \(BC\).

2. **To calculate:**
- We need to find the lengths of segments \(AB\), \(AH\), and also do the calculation for the height.

3. **Using the following formula to find \(AH\) (height from \(A\) to \(BC\)):**
\[
AH = \frac{AB \cdot BH}{BC}
\]
We need \(AB\) first for this step. We can calculate \(AB\) using the Pythagorean theorem once we gather values for the other sides.

4. **For \(DM \parallel AC\):**
- We need to apply similarity of triangles.

5. **For constructing point \(K\):**
- \(K\) is chosen so that \(AKM\) is right-angled at \(K\).

6. **Proofs needed:**
- Show that \(AE = BE\).
- Determine \(S^2_{AMK} = S_{AMB} \cdot S_{AMD}\).

To proceed, you'll need to express \(AB\), and then substitute back to find other segments, continually using properties of right triangles (such as similarity) and known segments until all required lengths are determined.

If you have specific lengths for \(AB\) or any other additional information, we can finalize calculations.