Richard sent us his work on this problem.
  1. We can say that A P S and B Q P are similar triangles (ratio of sides same), so in terms of angles:

    ÐA P S = ÐB Q P

    = 180° - 90° - ÐB P Q (angles in triangle add to 180°)

    =90° - ÐB P Q

    ÐS P Q + ÐA P S + ÐB P Q = 180° (angles from a point on a straight line add to 180°)

    Þ ÐS P Q + (90° - ÐB P Q) + ÐB P Q = 180°

    Þ ÐS P Q + 90° = 180°

    Þ ÐS P Q = 90°.

  2. Same applies again here, triangle Q C R is similar to R D S, and so in terms of angles:

    ÐC R Q = ÐD S R

    = 90° - ÐS R D

    ÐS R Q + ÐC R Q + ÐS R D = 180°

    ÐS R Q + (90° - ÐS R D) + ÐS R D = 180°

    ÐS R Q = 90°.

  3. As two opposite angles add up to 180°, the other two must as well (angles in quadrilateral add up to 360°). Two pairs of opposite angles each adding to 180° implies a cyclic quadrilateral (one of the Circle Theorems).
  4. Firstly, as ÐS P Q and ÐS R Q are both 90°, line S Q must be the diameter of the circle with midpoint being centre of circle, M.

    As A P R D is a rectangle, ÐA P R and ÐD R P are both right angles. Also lines P M = R M, as P and R are on the circle.

    Midpoint M is also the centre of the rectangle. Let V be the length of the perpendicular from M to side A B, and let U be the length from the foot of this perpendicular to P, as shown in the diagram.

    Rectangle with M, U and V added
    U = [y(x-y) + x(x+y)]/2 - A P

    = [x y - y2 + x2 + x y]/2 - y(x-y)

    = [x2 + y2]/2

    V = [y(x+y) + x(x-y)]/2

    = [x y + y2 + x2 -x y]/2

    = [x2 + y2]/2

    As U = V, ÐB P M = 45°

    Þ ÐR P M = 45°

    Þ ÐP R M = 45° (isosceles triangle)

    Þ ÐP M R = 180° - 45° - 45° = 90°

    Þ ÐP Q R = 45° (angle at centre is double that at edge - Circle Theorem)

  5. As ÐP Q R = 45°

    Þ ÐP S R = 180° - 45° (Cyclic Quadrilateral)

    = 135°.