Construct the $ \triangle ABC$, given $ h_a$, $ h_b$ (the altitudes from $ A$ and $ B$) and $ m_a$, the median from the vertex $ A$.
1. **Construct the altitude \( AH \) from vertex \( A \):**
   Given \( h_a \), we can construct the altitude \( AH \) such that \( AH = h_a \). Let \( H \) be the foot of the altitude from \( A \) to \( BC \).

2. **Construct the line \( d \) perpendicular to \( AH \) through \( H \):**
   Draw a line \( d \) that passes through \( H \) and is perpendicular to \( AH \).

3. **Construct the median \( AM \) from vertex \( A \):**
   Given \( m_a \), the median from \( A \) to \( BC \), we know that the median divides the opposite side into two equal segments. Let \( M \) be the midpoint of \( BC \). The point \( M \) lies on the line \( d \).

4. **Determine the ratio \( k \):**
   Since \( ah_a = bh_b \), we have:
   \[
   k = \frac{h_a}{h_b} = \frac{b}{a} = \frac{AC}{BC}
   \]
   This ratio \( k \) helps us in determining the relative lengths of the sides of the triangle.

5. **Use the Apollonius circle to find point \( C \):**
   The Apollonius circle of a line segment \( AM \) with respect to the ratio \( 2k \) is constructed. This circle will intersect the line \( d \) at point \( C \).

6. **Construct point \( B \):**
   Since \( M \) is the midpoint of \( BC \), we have:
   \[
   MC = MB = \frac{1}{2}BC
   \]
   Using this property, we can locate point \( B \) such that \( M \) is equidistant from \( B \) and \( C \).

By following these steps, we can construct the triangle \( \triangle ABC \) given the altitudes \( h_a \) and \( h_b \), and the median \( m_a \).

The final answer is the constructed triangle \( \boxed{ \triangle ABC } \).