Two flag poles of height $h$ and $k$ are situated $2a$ units apart on a level surface. Find the set of all points on the surface which are so situated that the angles of elevation of the tops of the poles are equal.
1. **Draw the Diagram and Define Variables:**
   - Let the two flag poles be \( A \) and \( B \) with heights \( h \) and \( k \) respectively.
   - The distance between the two poles is \( 2a \).
   - Let \( P \) be a point on the surface such that the angles of elevation to the tops of the poles are equal.
   - Let \( \theta \) be the common angle of elevation.
   - Let \( x \) be the distance from point \( P \) to the base of the pole of height \( k \).

2. **Set Up the Trigonometric Relationships:**
   - The angle of elevation to the top of the pole of height \( k \) is given by:
     \[
     \tan(\theta) = \frac{k}{x}
     \]
   - The distance from point \( P \) to the base of the pole of height \( h \) is \( 2a - x \).
   - The angle of elevation to the top of the pole of height \( h \) is given by:
     \[
     \tan(\theta) = \frac{h}{2a - x}
     \]

3. **Equate the Two Expressions for \(\tan(\theta)\):**
   \[
   \frac{k}{x} = \frac{h}{2a - x}
   \]

4. **Solve for \( x \):**
   - Cross-multiply to get:
     \[
     k(2a - x) = hx
     \]
   - Distribute \( k \):
     \[
     2ak - kx = hx
     \]
   - Combine like terms:
     \[
     2ak = hx + kx
     \]
     \[
     2ak = x(h + k)
     \]
   - Solve for \( x \):
     \[
     x = \frac{2ak}{h + k}
     \]

5. **Determine the Set of All Points:**
   - The distance \( x \) from the point \( P \) to the base of the pole of height \( k \) is \( \frac{2ak}{h + k} \).
   - The distance from the point \( P \) to the base of the pole of height \( h \) is \( 2a - x \):
     \[
     2a - x = 2a - \frac{2ak}{h + k} = \frac{2a(h + k) - 2ak}{h + k} = \frac{2ah}{h + k}
     \]

6. **Conclusion:**
   - The set of all points \( P \) on the surface such that the angles of elevation to the tops of the poles are equal is given by the distances \( x = \frac{2ak}{h + k} \) from the base of the pole of height \( k \) and \( 2a - x = \frac{2ah}{h + k} \) from the base of the pole of height \( h \).

The final answer is \( \boxed{ x = \frac{2ak}{h + k} } \) and \( 2a - x = \frac{2ah}{h + k} \).