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.**

**
**Proof.

**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

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