If P lies on the incircle, the result is trivial. So assume not.

Since the result deals only with tangency and concurrency, and there is a projective transformation fixing the incircle of ABC and taking P to the incentre, we may without loss of generality take P to be the incentre of ABC. But then by symmetry X lies at the midpoint of the arc EF so DX is the internal angle bisector of EDF. Similar statments hold for EY and FZ, so these three lines are concurrent at the incentre of DEF, as required.

You say:
'Line AP crosses the incircle in two points as do lines BP and CP. The applet picks the points nearest the vertices. However if X', Y', Z' are the farthest from the vertices points, then DX', EY' and FZ' are also concurrent. Both points of concurrency are known as Rabinowitz's points'

However, in the above case DX' is the external bisector of EDF, and similar results are true about EY' and FZ', so whilst the triplets (DX, EY', FZ'), (DX', EY, FZ') and (DX', EY', FZ) are each concurrent, the triplet (DX', EY', FZ') is not.

Thankyou

sfwc
<><

Thank you. That's good.

>
>Since the result deals only with tangency and concurrency,

That's good.

>However, in the above case DX' is the external bisector of
>EDF, and similar results are true about EY' and FZ', so
>whilst the triplets (DX, EY', FZ'), (DX', EY, FZ') and (DX',
>EY', FZ) are each concurrent, the triplet (DX', EY', FZ') is
>not.

Then which one is Rabinowitz's? (Just kidding.) I'll have to look into that.

Thank you