In a triangle $ABC$, let $H_a$, $H_b$ and $H_c$ denote the base points of the altitudes on the sides $BC$, $CA$ and $AB$, respectively.
Determine for which triangles $ABC$ two of the lengths $H_aH_b$, $H_bH_c$ and $H_aH_c$ are equal.
1. **Define the Orthic Triangle:**
   In a triangle \(ABC\), the orthic triangle is formed by the feet of the altitudes. Let \(H_a\), \(H_b\), and \(H_c\) be the feet of the altitudes from vertices \(A\), \(B\), and \(C\) to the opposite sides \(BC\), \(CA\), and \(AB\) respectively.

2. **Consider the Lengths \(H_aH_b\), \(H_bH_c\), and \(H_aH_c\):**
   We need to determine for which triangles \(ABC\) two of these lengths are equal. Let's assume \(H_aH_b = H_aH_c\).

3. **Use Properties of the Orthic Triangle:**
   The orthic triangle is similar to the original triangle \(ABC\). The angles of the orthic triangle are the same as the angles of the original triangle \(ABC\).

4. **Equal Lengths Imply Equal Angles:**
   If \(H_aH_b = H_aH_c\), then the corresponding angles in the orthic triangle must be equal. This means that the angles opposite these sides in the orthic triangle are equal. Therefore, \(\angle H_bH_aH_c = \angle H_cH_aH_b\).

5. **Relate to Original Triangle:**
   Since the orthic triangle is similar to the original triangle \(ABC\), the angles \(\angle H_bH_aH_c\) and \(\angle H_cH_aH_b\) correspond to the angles \(\angle B\) and \(\angle C\) in the original triangle \(ABC\). Therefore, \(\angle B = \angle C\).

6. **Conclusion:**
   For the lengths \(H_aH_b\) and \(H_aH_c\) to be equal, the triangle \(ABC\) must be isosceles with \(\angle B = \angle C\). The converse is also true: if \(\angle B = \angle C\), then \(H_aH_b = H_aH_c\).

The final answer is \( \boxed{ \angle B = \angle C } \).