First note that $\triangle{ABC}$ is the orthic triangle of $\triangle{I_AI_BI_C}$. Next, we show by angle chasing that $\triangle{I_ADE} \sim \triangle{I_ACB}$: $\angle{BCI_A} = 90-\frac{\angle{C}}{2} = 90-\left(90-\frac{\angle{B}}{2}-\frac{\angle{A}}{2}\right) = 90-\angle{AI_AB} = \angle{EDI_A}$ and likewise, $\angle{CBI_A} = \angle{DEI_A}$. Next, we show that $D, E$ are the feet of altitudes from $C, B$ in $\triangle{I_ACB}$: Using kite $PAQI_A$, $\frac{DI_A}{DI_C} = \tan^2\frac{\angle{A}}{2}$, and since $\angle{I_AI_CC} = \frac{\angle{A}}{2}$, D is foot of altitude from $C$. So $A_1$ is the orthocenter of $\triangle{I_ABC}$, and similarly for $B_1, C_1$. Thus $CA_1 \parallel BI, BA_1 \parallel CI \implies CA_1BI$ is a parallelogram and so are $BC_1AI, AB_1CI$, so $AA_1, BB_1, CC_1$ clearly all bisect each other and intersect at their common midpoint.