Diagonals of a quadrilateral $ABCD$ are equal and meet at point $O$. The perpendicular bisectors to segments $AB$ and $CD$ meet at point $P$, and the perpendicular bisectors to $BC$ and $AD$ meet at point $Q$. Find angle $\angle POQ$.

by A.Zaslavsky
1. **Congruence of Triangles:**
   - Given that the diagonals \(AC\) and \(BD\) of quadrilateral \(ABCD\) are equal and intersect at point \(O\), we can infer that \(\triangle AOB \cong \triangle COD\) by the SSS (Side-Side-Side) congruence criterion. This is because \(AO = OC\), \(BO = OD\), and \(AB = CD\) (since diagonals are equal and bisect each other).

2. **Concyclic Points:**
   - The perpendicular bisectors of segments \(AB\) and \(CD\) meet at point \(P\). Since \(P\) lies on the perpendicular bisectors of \(AB\) and \(CD\), it is equidistant from \(A\) and \(B\), and from \(C\) and \(D\). Therefore, \(PA = PB\) and \(PC = PD\).
   - Similarly, the perpendicular bisectors of segments \(BC\) and \(AD\) meet at point \(Q\). Thus, \(Q\) is equidistant from \(B\) and \(C\), and from \(A\) and \(D\). Therefore, \(QB = QC\) and \(QA = QD\).

3. **External Angle Bisectors:**
   - Since \(P\) is equidistant from \(A\) and \(B\), and from \(C\) and \(D\), \(P\) lies on the external angle bisector of \(\angle AOB\). This is because the external angle bisector of \(\angle AOB\) is the locus of points equidistant from \(A\) and \(B\).
   - Similarly, \(Q\) lies on the external angle bisector of \(\angle AOD\).

4. **Angle Between External Angle Bisectors:**
   - The external angle bisectors of \(\angle AOB\) and \(\angle AOD\) intersect at \(O\). Since \(\angle AOB\) and \(\angle AOD\) are supplementary (they add up to \(180^\circ\)), the angle between their external angle bisectors is \(90^\circ\).

Therefore, \(\angle POQ = 90^\circ\).

The final answer is \(\boxed{90^\circ}\)