Confusion on wording of an elementary geometry problem I really want to know the following geometry problem is valid or not. (Please don't change the wording of the problem. Please answer it is valid or not. Please answer frankly.)
"ABCD is a parallelogram. Any circle through A and B cuts DA and CB produced at P and Q respectively. Prove that DCQP is cyclic."
If the problem is valid, I also want to know the solution.
If the problem is not valid, I also want to know why.
Here the notation like XY means line segment XY.  If the points X, Y separate the line XY into three parts, the middle part (between X and Y) is the line segment XY. The part beyond Y (on the opposite side from X) is called XY produced. DCQP is a convex quadrilateral. 
 A: The statement is valid (to certain extend).

The figure drawn depends on (a) which angle is sharper, A or B; and which side is longer, AB or AD.
Apart from that, it also depends on where the center of the circle that passes through AB is.
(1) If it is at a location like O, we end up with the red circle cutting BC produced at Q. Then DCQP is cyclic. (The proof is not that difficult and is therefore skipped.)
(2) If it is at a location like K, we end up with the green circle cutting DA produced at P’ and CB produced at Q’. Then DCQ’P’ are cyclic. 
In Euclidean geometry, we have a very strict definition on the term ‘produced’. By saying Z lies on XY produced, we do mean Z is located closer to Y than X (and not the otherwise). Clearly, the statement has not considered such a distinction.
A: This problem is valid if you assume two points:


*

*DCQP can be a non-simple/ concave polygon.

*Interpret "DA and CB" as "line DA and line CB".


This is not considered change the wording of the question.
To solve it, first you see that a convex quadrilateral is cyclic $\iff$ the sum of two opposite angle of it is $180^\circ$.
Now, consider the case DCQP is convex.
Since ABCD is parallelogram, then $\angle DAB+\angle ADC= 180^\circ$.
Since Q lie on CB, then $\angle CQP+\angle PQB=180^\circ$.
ABQP is cyclic (by assumption) $\iff$ $\angle DAB+\angle PQB=180^\circ$.
From these you conclude $\angle ADC+\angle CQP=180^\circ\iff Q.E.D.$
Similar for the case DCQP is concave.
