Let a square ABCD with sides of length 1 be given. A point X on BC is at distance d
from C, and a point Y on CD is at distance d from C. The extensions of: AB and DX
meet at P, AD and BY meet at Q, AX and DC meet at R, and AY and BC meet at
S. If points P, Q, R and S are collinear, determine d.
I have started and found the diagram is symmetrical through AC, X must be (1,d) and Y must be (1-d,0). PR or QS must be at right angles to AC due to the need for a collinear (straight) line through them at the end. I believe the gradients should be the same for AC and QS, is this correct?? I'm stuck on what to do next and also need to clarify so far.