Mayank Kapoor, Campion School, Bhopal, India proved that the
  quadrilateral PQRS will always be a parallelogram. <br />
  <br />
  Considering first two intersecting circles with centres P and Q,
  XB is the common chord and PQ is the line joining the centres.
  Therefore XB is perpendicular to PQ. <br />
  <br />
  In the same way, XD is perpendicular to RS. XB and XD are on the
  same line and therefore PQ is parallel to RS. <br />
  <br />
  Proceeding in a similar manner and considering the circles with
  centres Q and R, we can prove that the line segments QR and PS
  are parallel. This completes the proof that PQRS is a
  parallelogram. <br />
  <br />
  Mayank's proof uses the fact that, for any two intersecting
  circles, the line joining the centres is perpendicular to the
  common chord. <br />
  <br />
  Yung Polam, from Hong Kong included a proof of this fact. <br />
  <br />
  Let BX intersect PQ at the point M. <br />
  <br />
  Consider triangles BPQ and XPQ. BQ = XQ (radii) and BP = XP
  (radii), PQ = PQ (common). Therefore triangles BPQ and XPQ are
  congruent. <br />
  <br />
  Therefore angle PQB = angle PQX But BQ = XQ (radii) and QM = QM
  (common) . Hence triangles BQM and XQM are congruent. <br />
  <br />
  Hence angle BMQ = angle XMQ (corr. angles, congruent triangles).
  Therefore angle XMQ = 90 degrees. <br />
  <br />
  Chi Kin Ho, St Dominic's International School of Lisbon, also
  gave an excellent proof including a discussion of the degenerate
  cases where one vertex, A, B, C or D is moved on top of another
  or on top of the point X. In these cases there is no longer a
  convex quadrilateral ABCD. For instance, if we move C on top of
  B, both the points B, C and X are joined at one point. As a
  result, the circumcircle of the triangle BXC is reduced to a
  point, and only 3 circles will be left. Quadrilateral PQRS cannot
  be formed any more. Consider also, as we move C on top of X, the
  common chord CX will be eventually reduced to a point, and RQ
  will therefore disappear. Quadrilateral PQRS can't be formed.
  Finally, if we move C on top of A, two of the circles will be
  overlapping the other two. As a result, RQ will also overlap PS,
  and the quadrilateral PQRS is reduced to a line. <br />