Let $ABC$ be an isosceles obtuse-angled triangle, and $D$ be a point on its base $AB$ such that $AD$ equals to the circumradius of triangle $BCD$. Find the value of $\angle ACD$.
1. **Identify the given information and the goal:**
   - $ABC$ is an isosceles obtuse-angled triangle with $AC = BC$.
   - $D$ is a point on the base $AB$ such that $AD$ equals the circumradius of triangle $BCD$.
   - We need to find the value of $\angle ACD$.

2. **Use the properties of the circumradius:**
   - The circumradius $R$ of a triangle with sides $a$, $b$, and $c$ and opposite angles $A$, $B$, and $C$ respectively is given by:
     \[
     R = \frac{a}{2 \sin A}
     \]
   - In this problem, $AD = R_{BCD}$, where $R_{BCD}$ is the circumradius of triangle $BCD$.

3. **Apply the Sine Rule in $\triangle BCD$:**
   - By the Sine Rule in $\triangle BCD$, we have:
     \[
     \frac{BC}{\sin \angle BDC} = 2R_{BCD}
     \]
   - Since $BC = AC$, we can write:
     \[
     \frac{AC}{\sin \angle BDC} = 2R_{BCD}
     \]

4. **Relate $AD$ to $R_{BCD}$:**
   - Given $AD = R_{BCD}$, we substitute $R_{BCD}$:
     \[
     \frac{AC}{\sin \angle BDC} = 2AD
     \]

5. **Use the Sine Rule in $\triangle ACD$:**
   - By the Sine Rule in $\triangle ACD$, we have:
     \[
     \frac{AD}{\sin \angle ACD} = \frac{AC}{\sin \angle ADC}
     \]
   - Since $AD = R_{BCD}$ and $R_{BCD} = \frac{AC}{2 \sin \angle BDC}$, we substitute:
     \[
     \frac{AD}{\sin \angle ACD} = 2AD
     \]

6. **Simplify the equation:**
   - Dividing both sides by $AD$, we get:
     \[
     \frac{1}{\sin \angle ACD} = 2
     \]
   - Therefore:
     \[
     \sin \angle ACD = \frac{1}{2}
     \]

7. **Determine $\angle ACD$:**
   - The angle whose sine is $\frac{1}{2}$ is $30^\circ$:
     \[
     \angle ACD = 30^\circ
     \]

The final answer is $\boxed{30^\circ}$