Of a rhombus $ABCD$ we know the circumradius $R$ of $\Delta ABC$ and $r$ of $\Delta BCD$. Construct the rhombus.
To construct the rhombus \(ABCD\) given the circumradius \(R\) of \(\Delta ABC\) and the inradius \(r\) of \(\Delta BCD\), we can follow these steps:

1. **Identify the centers \(O\) and \(I\)**:
   - Let \(O\) be the circumcenter of \(\Delta ABC\) with radius \(R\).
   - Let \(I\) be the incenter of \(\Delta BCD\) with radius \(r\).

2. **Establish the perpendicularity**:
   - Since \(I\) is on \(AC\), the perpendicular bisector of \(BD\), and \(O\) is on \(BD\), the perpendicular bisector of \(AC\), we have \(OI \perp BC\).

3. **Determine the orthocenter**:
   - \(I\) is the orthocenter of \(\Delta OBC\).

4. **Reflect \(I\) across \(BC\)**:
   - Let \(J\) be the reflection of \(I\) across \(BC\). Then, \(OBCJ\) is a cyclic quadrilateral.

5. **Construct the circle \((OBCJ)\)**:
   - The center of the circle \((OBCJ)\) lies on the line \(OJ\) because \(\Delta OBC\) and \(\Delta JBC\) are isosceles.

6. **Perpendicularity of \(OC\) and \(CJ\)**:
   - Since \(OBC\) and \(JBC\) are isosceles, \(OC\) and \(CJ\) are perpendicular.

Now, we can begin the construction:

1. **Draw the circle \((O)\)**:
   - Draw a circle with center \(O\) and radius \(R\). Place point \(C\) on this circle.

2. **Draw segment \(CJ\)**:
   - Draw a segment \(CJ\) perpendicular to \(OC\).

3. **Construct the circle \((OCJ)\)**:
   - Draw the circle passing through points \(O\), \(C\), and \(J\).

4. **Find point \(B\)**:
   - The point \(B\) is the second intersection of the circles \((O)\) and \((OCJ)\).

5. **Draw line \(BO\)**:
   - Draw the line \(BO\).

6. **Find point \(A\)**:
   - Draw the perpendicular to \(BO\) through \(C\). The intersection of this perpendicular with the circle \((O)\) gives point \(A\).

7. **Find point \(D\)**:
   - The symmetric point of \(B\) with respect to \(AC\) is point \(D\).

Thus, we have constructed the rhombus \(ABCD\).

The final answer is the constructed rhombus \( \boxed{ ABCD } \).