In a non-isosceles triangle $ABC$ the bisectors of angles $A$ and $B$ are inversely proportional to the respective sidelengths. Find angle $C$.
1. Let \( N \) and \( M \) be the feet of the altitudes from \( A \) and \( B \), respectively. We know that the angle bisectors \( AD \) and \( BE \) are inversely proportional to the respective side lengths \( BC \) and \( AC \). This implies:
   \[
   AD \cdot BC = BE \cdot AC
   \]
   and
   \[
   AN \cdot BC = BM \cdot AC
   \]
   Therefore,
   \[
   \frac{AD}{AN} = \frac{BE}{BM}
   \]
   This implies that the triangles \( \triangle ADE \) and \( \triangle BME \) are similar, and thus:
   \[
   \angle MEB = \angle NDA
   \]

2. We consider two cases:
   - **Case 1:** \( AM - AE \) and \( BD - BN \) have the same sign. Since \( \angle MEB = \angle NDA \), points \( A \), \( E \), \( D \), and \( B \) are concyclic. Because \( AD \) and \( BE \) are angle bisectors, we have:
     \[
     \angle BAD = \angle EAD = \angle EBD = \angle EBA
     \]
     This implies:
     \[
     \angle A = \angle B
     \]
     which contradicts the non-isosceles hypothesis.

   - **Case 2:** \( AM - AE \) and \( BD - BN \) have different signs. Since \( \angle MEB = \angle NDA \), points \( C \), \( E \), \( I \), and \( D \) are concyclic. Let:
     \[
     \angle IAB = \alpha \quad \text{and} \quad \angle IBA = \beta
     \]
     Then:
     \[
     \angle EIA = \angle DIB = \alpha + \beta
     \]
     Therefore:
     \[
     \angle CEI = \angle EAI + \angle EIA = 2\alpha + \beta
     \]
     and similarly:
     \[
     \angle CDI = \alpha + 2\beta
     \]
     Since \( CEID \) is cyclic, we have:
     \[
     \angle CEI + \angle CDI = 180^\circ
     \]
     Thus:
     \[
     2\alpha + \beta + \alpha + 2\beta = 180^\circ
     \]
     Simplifying, we get:
     \[
     3\alpha + 3\beta = 180^\circ \quad \Rightarrow \quad \alpha + \beta = 60^\circ
     \]
     Therefore:
     \[
     \angle C = 180^\circ - 2\alpha - 2\beta = 180^\circ - 2(\alpha + \beta) = 180^\circ - 2 \cdot 60^\circ = 60^\circ
     \]

The final answer is \(\boxed{60^\circ}\)