Just to pass the time, a very simple result (again found by playing on Geogebra):
A triangle ABC is acute-angled. The point P is located in the interior of ABC. The point Q is located on the interior of BC. The circumcircles of PBQ and PQC meet AB and AC again at R and S respectively. Then:
R, B, C and S are concyclic if and only if A, P and Q are collinear.
1. If A, P and Q are collinear, then RBCS is a cyclic quadrilateral.
The first thing to notice is that ARPS is also a cyclic quadrilateral. (This happens regardless of whether A, P and Q are collinear or not.) The proof of this is just angle chasing; this is in fact a direct consequence of Miquel’s theorem, for those of you who have heard of it. (I think it’s underrated.)
Let ∠PSC = x
∠ASP = 180 – x (straight line)
∠PQC = 180 – x (opposite angle in cyclic quadrilateral PQCS)
∠PQB = x (straight line)
∠PRB = 180 – x (opposite angle in cyclic quadrilateral RBQP)
∠ARP = x (straight line)
∠ARP + ∠ASP = 180 so ARPS is cyclic by the converse of ‘opposite angles in a cyclic quad are supplementary’
Now let ∠RSP = y
∠RAP = y (same segment)
As ∠PQB = x, we can use angles in a triangle in ABQ to obtain ∠RBC = 180 – x – y (this is allowed as APQ is a straight line)
Then ∠RBC + ∠RSC = 180 so RBCS is cyclic as opposite angles are supplementary.
2. If RBCS is a cyclic quadrilateral, then A, P and Q are collinear.
Again let ∠PSC = x and ∠RSP = y.
∠ASR = 180 – x – y (straight line)
∠APR = 180 – x – y (same segment)
Also ∠RBC = 180 – x – y (opposite angle in cyclic quadrilateral RBCS)
∠RPQ = x + y (opposite angle in cyclic quadrilateral RBQP)
As ∠RPQ + ∠APR = 180, APQ is a straight line.
There it is