Geometry in the circle.?

@ Detective. Thx for the pic!

@ Sharmistha: P,Q,R are arbitrary and S is determined by the condition SQ' || PR. So in general PQR is not a right angle. If it is then S = Q' completes the rectangle and the question is trivial.

@ AI P. Could not agree more with what you said but the connection to this q totally eludes me.

@ Sharmistha: QQ' is a diameter only if PR is a diameter. In this case PQRS is a kite and P = R'. So the line PR' becomes the tangent at P which is indeed parallel to QS. But again, that's not the general case.

@ Sharmistha: there is no way to dispense with the middles of the diagonals.

Besides your alternate statement involving angular bissectors appears to be false: indeed if QQ' is the angular bissector of angle Q, that simply means that Q' is the middle of the arc limited by P and R not containing Q. So the tangent at Q' is parallel to PR and this forces Q' = S. But in general P is not the middle of arc(QS) so that P and R' are distinct and PR' is not parallel to QS.

@ AI P: pretty pic! PR' and QS appear to be at a slight angle... any idea where this comes from?

@ Sharmistha: remember that PQR hence QQ' are totally arbitrary. Therefore what you say implies that any bissector goes through the center, hence is a diameter. Or am I misreading you?

@ AI P "Let PQRS a quadrilateral inscribed in a circle C." So all vertices are on C.

@ AI P Good luck with your pics, but what do you hope to find when they are not all the same circle?

@ Sharmistha: I indicated that this question is a rewording. If you look at the original asker's questions, you'll see that they are challenging but not daunting. So I would not go for anything too complicated.

@ AI P. Yes P,Q,R,S,Q' and R' are all on the same euclidean circle. P,Q,R are arbitrary and S,Q',R' are then determined using the assumptions.

@ Sharmistha: "If the contention is correct that the "special" corner of the cyclic q when joined with the mid point of the line joining it's two adjacent corners (the other diagonal of the q) bisects the angle at the special corner" Just stop right here. This not at all the contention. Anyway at this stage there is only a definition which is that this line and QS are symmetric with respect to the angle bissector at Q, which is equivalent to saying that SQ' || PR.

In the initial problem, the angle bissector at B, is also the angle bissector between BI and BD.

The pics of AI P and Detective give ample evidence of the truth of the statement, so I see no need to reword anything, Making your own pic is all I can suggest... good luck!

@ Zo Maar: Very good. You even correctly guessed that I was not crazy about brute force!

@ AI P: I'll leave it open till expiration time minus one hour.

@ Sharmistha: Yes your first hypothesis is correct. It follows from

http://en.wikipedia.org/wiki/Inscribed_angle#Theor...

This theorem says that equal angles intersect arcs of the same length. So the bissector at Q intersect the circle at a point M which is on the mediatrix of PR. Similarly the arcs SM and MQ' have the same length, which implies that SQ' || PR. This is the fact which suggested the rewording of the original question.

As for your contestation, I suggest you try to find a flaw in Zo Maar's solution!

@ Sharmistha: "arc SM and MQ' are of the same length but they are not on the same straight line."

Of course arcs are not on a line. " Hence it's not a reflection of point S on point Q' over the angular bissector." Right it's a reflection around the mediatrix of PR.

The original question talks of lines QS and QQ' symmetric with respect to the angular bissector at Q, but in my previous post I explained why it is equivalent to the present formulation.

Don't be afraid: my deduction is correct and we already have 1 proof and 2 pics which support the question as stated.

@ Indica: Beautiful. I guess the most resilient answerer should be convinced!