Problem 1610

1. Identify the given elements and relationships:
- We have an inscribed quadrilateral $ABCD$ with midpoints $M, N, P, Q$ of sides $AB, BC, CD, DA$ respectively.
- Diagonals $AC$ and $BD$ intersect at point $O$.

2. Establish similarity of triangles:
- Since $M$ and $P$ are midpoints, $BM = \frac{AB}{2}$ and $PC = \frac{CD}{2}$.
- Given $\widehat{MBO} \equiv \widehat{PCO}$ and $\frac{OB}{OC} = \frac{AB}{CD} = \frac{BM}{PC}$, we can conclude that $\triangle BOM \sim \triangle COP$.

3. Use the Law of Sines in similar triangles:
- From the similarity $\triangle BOM \sim \triangle COP$, we have:
$$\frac{OM}{OP} = \frac{BO}{OC} = \frac{\sin \angle ACB}{\sin \angle DBC}$$
- Applying the Law of Sines in $\triangle BOM$ and $\triangle COP$:
$$\frac{OM}{\sin \angle ACB} = \frac{OP}{\sin \angle DBC}$$

4. Relate areas of triangles:
- Using the area formula for triangles, we have:
$$\frac{ON \cdot OM \cdot MN}{BC \cdot AC \cdot \sin \angle ACB} = \frac{ON \cdot OP \cdot NP}{BC \cdot BD \cdot \sin \angle DBC}$$
- Simplifying, we get:
$$\frac{ON \cdot OM \cdot MN}{S_{\triangle ACB}} = \frac{ON \cdot OP \cdot NP}{S_{\triangle DBC}}$$

5. Equate the areas:
- It follows that:
$$4S_{\triangle MON} = S_{\triangle ACB} \quad \text{and} \quad 4S_{\triangle PON} = S_{\triangle DBC}$$
- Therefore:
$$\frac{ON \cdot OM \cdot MN}{S_{\triangle MON}} = \frac{ON \cdot OP \cdot NP}{S_{\triangle PON}}$$
- This implies:
$$R_{\triangle MON} = R_{\triangle PON}$$

6. Generalize for other triangles:
- By similar arguments, we can prove the same relationship for $\triangle ONP$ and $\triangle OQM$.

Conclusion:
$$\boxed{R_{\triangle MON} = R_{\triangle PON} = R_{\triangle ONP} = R_{\triangle OQM}}$$