<Tri-2025-2268>, hint 

Let M ,N be the midpoints of AC, AB,respectively

Suppose tangent at F meet AB at P' and CP is the bisector of angle C, P on AB.

By angle chasing , <AMN=<C, <IP'F=<(3/2)C, IF=r (inradius), <IFP'=90

Incircle touches AB at Q, <CPQ=<(3/2)C. Thus triangles IFP' and IQP are congruent

Hence P=P'