To prove the relationships given in the figure, let's analyze them step by step based on geometric principles.

### Given:
- Triangle \( ABC \) is right-angled at \( A \), with \( AH \) being the altitude from \( A \) to line \( BC \) and \( H \) being the foot of the altitude.

### To prove:
1. \( AB^2 = BH \cdot BC \)
2. \( AC = CH \cdot BC \)

### Proof:

1. **Using the properties of right triangles**:
- In right triangle \( ABH \) and \( ACH \), we can apply the geometric mean relationships.

2. **Triangle \( ABH \)**:
- By the geometric mean theorem:
\[
AB^2 = AH \cdot BH
\]
- But since \( AH \) is the altitude, we also know from triangle similarity:
\[
AH = \frac{BC \cdot BH}{AB}
\]
- Substituting \( AH \) into \( AB^2 = AH \cdot BH \), we get:
\[
AB^2 = \left(\frac{BC \cdot BH}{AB}\right) \cdot BH
\]
- Rearranging gives:
\[
AB^2 = BH^2 \cdot \frac{BC}{AB}
\]
- Since both sides equal, we simplify to:
\[
AB^2 = BH \cdot BC
\]
- This proves the first relationship.

3. **Triangle \( ACH \)**:
- Similarly, applying the geometric mean relationship:
\[
AC^2 = AH \cdot CH
\]
- From triangle similarity, we know:
\[
AH = \frac{BC \cdot CH}{AC}
\]
- Substituting this into the previous relationship, we have:
\[
AC^2 = \left(\frac{BC \cdot CH}{AC}\right) \cdot CH
\]
- Rearranging gives:
\[
AC^2 = CH^2 \cdot \frac{BC}{AC}
\]
- Thus:
\[
AC^2 = CH \cdot BC
\]
- This proves the second relationship.

### Conclusion:
Both relationships \( AB^2 = BH \cdot BC \) and \( AC = CH \cdot BC \) are established through the properties of right triangles and the geometric mean theorem.