The incircle of $ABC$ touches the sides $BC,CA,AB$ at $A' ,B' ,C'$ respectively. The line $A' C'$  meets the angle bisector of $\angle A$ at $D$. Find $\angle ADC$.
1. **Identify the given elements and their properties:**
   - The incircle of $\triangle ABC$ touches the sides $BC$, $CA$, and $AB$ at $A'$, $B'$, and $C'$ respectively.
   - The line $A'C'$ meets the angle bisector of $\angle A$ at $D$.
   - We need to find $\angle ADC$.

2. **Use the property of the incircle and cyclic quadrilaterals:**
   - Since $A'$, $B'$, and $C'$ are points where the incircle touches the sides, $A'$, $B'$, and $C'$ are known as the points of tangency.
   - The quadrilateral $BC'IA'$ is cyclic because the opposite angles sum up to $180^\circ$.

3. **Calculate the angles using the properties of cyclic quadrilaterals:**
   - Since $BC'IA'$ is cyclic, we have:
     \[
     \angle C'A'I = \frac{\angle B}{2}
     \]
   - Notice that $\angle DA'C = 90^\circ + \frac{\angle B}{2}$ because $A'$ is the point of tangency and $D$ lies on the angle bisector of $\angle A$.

4. **Use the properties of the incenter and angle bisectors:**
   - The incenter $I$ is the intersection of the angle bisectors of $\triangle ABC$.
   - The angle $\angle BID$ is given by:
     \[
     \angle BID = \frac{1}{2} (\angle A + \angle B)
     \]
   - The angle $\angle BIC$ is:
     \[
     \angle BIC = 90^\circ + \frac{\angle A}{2}
     \]

5. **Calculate $\angle DIC$:**
   - Using the angles calculated above:
     \[
     \angle DIC = \angle BIC - \angle BID = \left(90^\circ + \frac{\angle A}{2}\right) - \left(\frac{1}{2} (\angle A + \angle B)\right)
     \]
     Simplifying this:
     \[
     \angle DIC = 90^\circ + \frac{\angle A}{2} - \frac{\angle A}{2} - \frac{\angle B}{2} = 90^\circ - \frac{\angle B}{2}
     \]

6. **Conclude that $A'DIC$ is cyclic:**
   - Since $\angle DIC = 90^\circ - \frac{\angle B}{2}$, we can conclude that $A'DIC$ is cyclic.

7. **Determine $\angle ADC$:**
   - In a cyclic quadrilateral, the opposite angles sum up to $180^\circ$.
   - Therefore, $\angle ADC = 180^\circ - \angle A'DI$.
   - Since $\angle A'DI = 90^\circ$, we have:
     \[
     \angle ADC = 180^\circ - 90^\circ = 90^\circ
     \]

The final answer is $\boxed{90^\circ}$