Who can solve it?

let $ D_1=AB\cap (I),D_2=BC\cap (I),K=DD_1\cap EG,L=DD_2\cap EF$ claim $EK=EL$ : we have $D_1KE\sim D_1ED$ so $\frac{KE}{ED}=\frac{D_1E}{D_1D}$ similarly we get $\frac{LE}{ED}=\frac{D_2E}{D_2D}$ but $EDD_1D_2$ is harmonic thus $\frac{D_1E}{D_1D}=\frac{D_2E}{D_2D}$ hence the claim follows and $KL $ is perpendicular to the bisector of $\angle ABC$. $\frac{KI_2}{I_2D}=\frac{KE}{ED}=\frac{LE}{ED}=\frac{LI_1}{I_1D} $ whence $I_1I_2 \parallel KL$ so the result required follows. RH HAS