Let $A,B,C$ be points on the sides $B_1C_1, C_1A_1,A_1B_1$ of a triangle $A_1B_1C_1$ such that $A_1A,B_1B,C_1C$ are the bisectors of angles of the triangle. We have that $AC = BC$ and $A_1C_1 \neq B_1C_1.$
$(a)$ Prove that $C_1$ lies on the circumcircle of the triangle $ABC$.
$(b)$ Suppose that $\angle BAC_1 =\frac{\pi}{6};$ find the form of triangle $ABC$.
### Part (a)

1. **Given Conditions and Setup:**
   - Points \( A, B, C \) lie on the sides \( B_1C_1, C_1A_1, A_1B_1 \) of triangle \( A_1B_1C_1 \).
   - \( A_1A, B_1B, C_1C \) are the angle bisectors of \( \triangle A_1B_1C_1 \).
   - \( AC = BC \) and \( A_1C_1 \neq B_1C_1 \).

2. **Angle Bisector Property:**
   - Since \( A_1A, B_1B, C_1C \) are angle bisectors, they intersect at the incenter \( I \) of \( \triangle A_1B_1C_1 \).

3. **Isosceles Triangle Property:**
   - Given \( AC = BC \), triangle \( \triangle ABC \) is isosceles with \( AC = BC \).

4. **Angle Equality:**
   - Since \( A_1A, B_1B, C_1C \) are angle bisectors, we have \( \angle AC_1C = \angle BC_1C \).

5. **Cyclic Quadrilateral Condition:**
   - For quadrilateral \( C_1ACB \) to be cyclic, the opposite angles must sum to \( 180^\circ \). Specifically, we need \( \angle ACB + \angle AC_1B = 180^\circ \).

6. **Contradiction Argument:**
   - If \( C_1A = C_1B \), then \( \triangle C_1AB \) would be isosceles with \( C_1A = C_1B \). This would imply \( A_1C_1 = B_1C_1 \), contradicting the given \( A_1C_1 \neq B_1C_1 \).

7. **Conclusion:**
   - Therefore, \( C_1 \) must lie on the circumcircle of \( \triangle ABC \).

\(\blacksquare\)

### Part (b)

1. **Given Condition:**
   - \( \angle BAC_1 = \frac{\pi}{6} \).

2. **Using the Cyclic Property:**
   - Since \( C_1 \) lies on the circumcircle of \( \triangle ABC \), we can use the properties of cyclic quadrilaterals.

3. **Isosceles Triangle Property:**
   - Given \( AC = BC \), \( \triangle ABC \) is isosceles with \( AC = BC \).

4. **Angle Calculation:**
   - In \( \triangle ABC \), let \( \angle BAC = \alpha \). Since \( \triangle ABC \) is isosceles, \( \angle ABC = \angle ACB = \beta \).

5. **Sum of Angles in Triangle:**
   - The sum of angles in \( \triangle ABC \) is \( 180^\circ \):
     \[
     \alpha + 2\beta = 180^\circ
     \]

6. **Using Given Angle:**
   - Given \( \angle BAC_1 = \frac{\pi}{6} = 30^\circ \), and since \( C_1 \) lies on the circumcircle, \( \angle BAC_1 = \angle BAC \):
     \[
     \alpha = 30^\circ
     \]

7. **Solving for \( \beta \):**
   - Substitute \( \alpha = 30^\circ \) into the angle sum equation:
     \[
     30^\circ + 2\beta = 180^\circ
     \]
     \[
     2\beta = 150^\circ
     \]
     \[
     \beta = 75^\circ
     \]

8. **Conclusion:**
   - Therefore, the angles of \( \triangle ABC \) are \( 30^\circ, 75^\circ, 75^\circ \).

\(\boxed{30^\circ, 75^\circ, 75^\circ}\)