Something else I stumbled across during my ramblings on GeoGebra:

Extend each side of the triangle ABC in both directions. The point P lies on BC; the circle ABP cuts AC again at S; the circle ACP cuts AB again at T. Circles BPT and CPS intersect at P and at X.

**A) As P moves along the line BC, X moves around a fixed circle.
B) Also, this fixed circle has the same radius as the circle ABC.
**This holds whether ABC is acute, right angled or obtuse; P is also not constrained to be between B and C, but can lie anywhere on the line – similarly, the points T and S do not have to lie on the interiors of AB and AC.

Here is a diagram of when P is not located on the interior of BC:

The green circle, along which X moves, has not changed.

**Proof of A)
Case 1
**First we will show that X is constrained to lie on a fixed circle when P is located between B and C.

Let ∠ABC = b and ∠ACB = c

By opposite angles in the cyclic quad BPSA, ∠PSA = 180 – b

∠PSC = b (straight line)

∠PXC = b (same segment in cyclic quad PCSX)

Similarly, by opposite angles in the cyclic quad PCAT, ∠PTA = 180 – c

∠PTB = c (straight line)

∠PXB = c (same segment in cyclic quad BPXT)

So ∠BXC = b + c

This is independent of the position of P; therefore, for all P on the interior of BC, the angles ∠BXC are all equal, and hence all the X’s lie on a common circle that passes through B and C (by the converse of the same segment theorem).

*As a side-note, we can deduce that BXS and CXT are straight lines; ∠PXB = c, and ∠PXS = 180 – c (opposite angles in cyclic quadrilateral PCSX); these add to 180 so P, X and S are collinear.*

*Similarly, C, X and T are collinear.*

**Case 2***
*Now we will show that when P is not between B and C, X still moves along a fixed circle, and also that this fixed circle is the same circle as in Case 1.

Using same segment theorem in cyclic quadrilateral ABSP, ∠ABP = ∠ASP = b

Using same segment theorem in cyclic quadrilateral CXSP, ∠CSP = ∠CXP = b

Also, ∠ACP = 180 – c (straight line)

∠ATP = c (opposite angle in cyclic quadrilateral ACPT)

∠BXP = 180 – c (opposite angle in cyclic quadrilateral BXPT)

So ∠BXC = ∠BXP – ∠CXP = 180 – b – c

Again, this is independent of the location of P, so X moves along a fixed circle. Also, this angle is 180 minus the angle BXC in Case 1. Therefore, by the converse of opposite angles in a cyclic quadrilateral are supplementary, this circle is the same circle as the circle in Case 1.

*Note that X will never completely trace round the circle; as P travels along the line BC to infinity, X will approach, but never meet, the intersection of the circle and the perpendicular bisector of BC.*

**Proof of B)**

This is very quick. Refer to the diagram from B) case 1:

**
**We employ the

*full*sine rule:

where R is the radius of the circumcircle.

In triangle ABC, the diameter of the circumcircle is:

In triangle XBC, the diameter of the circumcircle is:

But ∠BAC = ∠BXC = 180 – b – c

Hence the lengths of the diameters and radii of circles ABC and XBC are equal.