In the figure, $C$ is the point of contact of incircle of $\triangle DEF $ and $DE$. $I_1$ is the incentre of $\triangle DCF$ and $I_1G \perp FC$. $I_2$ is the incentre of $\triangle CEF$. Prove that $I_2G \perp CF$.
$I_1G \perp FC \implies I_1G$ is the inradius of incircle of $\triangle DCF$. I tried to do 'angle chasing' but I am not able to use line $CF$. I think finding $\angle FCE$ in terms of vertex angles of $\triangle DEF$ would work, but I am not able to figure out any such relation. Please help!
Since $I_1$ is the incenter, we have
$$2CG = FC + DC - FD\tag{1}$$
Also, given $C$ is the contact point of incircle $\triangle DEF$, we have
$$FD - FE = DC - EC\tag{2}$$
Plug (2) into (1), we get
$$2CG = FC + EC - FE\tag{3}$$
which means, just as for the incircle with center $I_1$, $G$ is also the tangential point for the incircle of center $I_2$. (Assume the tangential point $X$, $CX=CG$ from (3). Hence, $G$ and $X$ are the same point.)
Thus,
$$I_2G \perp CF$$