In triangle $ABC$ we have $|AB| \ne |AC|$. The bisectors of $\angle ABC$ and $\angle  ACB$ meet $AC$ and $AB$ at $E$ and $F$, respectively, and intersect at I. If $|EI| = |FI|$ find the measure of $\angle  BAC$.
1. **Applying the Angle Bisector Theorem:**
   By the Angle Bisector Theorem, we know that the angle bisectors of a triangle divide the opposite sides in the ratio of the adjacent sides. Therefore, for $\triangle ABC$:
   \[
   \frac{BE}{EC} = \frac{AB}{AC} \quad \text{and} \quad \frac{CF}{FB} = \frac{AC}{AB}
   \]

2. **Using the given condition $|EI| = |FI|$:**
   Since $I$ is the incenter of $\triangle ABC$ and $|EI| = |FI|$, it implies that $I$ is equidistant from $E$ and $F$. This means that $I$ lies on the perpendicular bisector of $EF$.

3. **Establishing the relationship between the segments:**
   From the Angle Bisector Theorem, we have:
   \[
   \frac{BI}{IE} = \frac{BC}{CE} \quad \text{and} \quad \frac{CI}{IF} = \frac{BC}{BF}
   \]
   Given $|EI| = |FI|$, we can write:
   \[
   BC \cdot IE = BI \cdot CE \quad \text{and} \quad BC \cdot IF = CI \cdot BF
   \]
   Since $IE = IF$, it follows that:
   \[
   \frac{BI}{BF} = \frac{CI}{CE}
   \]

4. **Considering the angles:**
   Let $\angle BFC = x$, $\angle CEB = y$, and $\angle BIF = \angle CIE = a$. By the Law of Sines in $\triangle BIF$ and $\triangle CIE$, we have:
   \[
   \frac{\sin x}{\sin a} = \frac{\sin y}{\sin a}
   \]
   This implies:
   \[
   \sin x = \sin y
   \]
   Therefore, $x = y$ or $x = 180^\circ - y$.

5. **Analyzing the angle relationships:**
   Since $|AB| \ne |AC|$, we must have $x + y = 180^\circ$. This implies that $A, F, I, E$ are concyclic (i.e., they lie on the same circle).

6. **Using the known cyclic property:**
   It is well known that if $A, F, I, E$ are concyclic, then:
   \[
   \angle FIE = 90^\circ + \frac{1}{2} \angle A
   \]
   Given the cyclicity, we have:
   \[
   \angle BAC = 60^\circ
   \]

The final answer is $\boxed{60^\circ}$.