Just notice that $\angle IFA = \angle IAC = \angle IAF \rightarrow IA = IF$. $I$ is on the perpendicular bisector of $AD$, so $ID = IF$. We have now that $I$ is the circumcenter of $ADF$, so $BI$ is the perpendicular bisector of $FD$, and bisector of $\angle BDF$, from this $BD = BF$, as desired.

Solution. First off, the tangency condition of $AC$ and $\Gamma$ passing through $I$ translates into $IA=IF$. Moreover, $I$ lies on the perpendicular bisector of $\overline{AD}$, thus $I$ is the circumcenter of $\bigtriangleup AFD$. We have $$\angle DFA = \angle DFI+\angle AFI = \angle FDI+\angle IAC=\angle FDI+\angle IDC = \angle CDF$$which readily leads to the required assertion. $\square$

Approach 1. After proving $IA=IF$, it is easy to check that $\angle IAD = \dfrac{\beta}{2}$ thus $BDIA$ is cyclic, and since $BI$ is bisector of $\angle ABD$ it yields $IA=ID$, now $\angle BFI = 180^{\circ} - \angle AFI = 180^{\circ} - \angle FAI = \angle BDI$, finally from this we have $\triangle BFI \equiv \triangle BDI$ and we are done. Approach 2. A metric approach would be that $BD=b-a$ and after proving $IA=IF$ we take $E$ the projection of $I$ onto $AB$ such that $BF=BE-FE=BE-AE=(s-b)-(s-a)=b-a$, finally $BD=BF$

This was proposed by Martín Díaz from Colombia