Perpendicularities $AD \perp BC$ and $BE \perp AC$ are unnecessary; we only need points $D,E$ on $BC,AC,$ such that $AEDB$ is cyclic and $AE+BD=BC.$ Using the converse of 9th ibmo - brazil 1994/q2 (i), it follows that the circle with center $F$ tangent to $CB,CA$ also touches $DE$ $\Longrightarrow$ $F$ becomes C-excenter of $\triangle CDE$ $\Longrightarrow$ $D,E$ are on circle with diameter $\overline{FI_C}.$ Angle chase gives $\widehat{DI_BF}=90^{\circ}+\tfrac{1}{2}\widehat{DBA}=90^{\circ}+\tfrac{1}{2}(180^{\circ}-\widehat{DEA})=180^{\circ}-\widehat{DEF}$ $\Longrightarrow$ $I_B \in \odot(DEF).$ Likewise $I_A \in \odot(DEF),$ hence $D,E,F,I_A,I_B,I_C$ are concyclic.