To solve the problem, let's analyze the information given.

**Triangle \( ABC \)** is isosceles with \( AB = AC = 5 \, \text{cm} \). The altitude \( AH \) is given as \( 4 \, \text{cm} \) and it is perpendicular to \( BC \).

### Part a: Prove that \( HB = HC \)

Since triangle \( ABC \) is isosceles and \( AH \) is the altitude from \( A \) to \( BC \), it bisects \( BC \). Thus, we have \( HB = HC \).

### Part b: Prove that \( AH \) is the median of triangle \( ABC \)

Since \( AH \) is perpendicular to \( BC \) and bisects \( BC \), it serves as both an altitude and a median. Therefore, \( AH \) is indeed the median of triangle \( ABC \).

### Part c: Calculate the perimeter of triangle \( ABC \)

To find the length of \( BC \), we can use the Pythagorean theorem in triangle \( AHB \):

\[
AB^2 = AH^2 + HB^2
\]
Substituting the values:
\[
5^2 = 4^2 + HB^2
\]
\[
25 = 16 + HB^2
\]
\[
HB^2 = 9 \implies HB = 3 \, \text{cm}
\]

Since \( HB = HC \), it follows that \( HC = 3 \, \text{cm} \).

Now, we can calculate \( BC \):
\[
BC = HB + HC = 3 + 3 = 6 \, \text{cm}
\]

Finally, the perimeter \( P \) of triangle \( ABC \) is given by:
\[
P = AB + AC + BC = 5 + 5 + 6 = 16 \, \text{cm}
\]

### Conclusion

- a) \( HB = HC \)
- b) \( AH \) is a median.
- c) The perimeter of triangle \( ABC \) is \( 16 \, \text{cm} \).