Similar triangles yield the solution due to $AI_a.AI=AB.AC$ and $AA_1.2R=AB.AC$

From angle chasing we get that $\measuredangle IBI_A=90^{\circ}$ , but we have $\measuredangle CBI=\measuredangle IBA$ , now let $AI \cap BC=L$ and $AI \cap \omega=\{A,S\}$ ,where $\omega$ is the circumcircle of the triangle $ABC$. Therefore we get $(A,L;I,I_A)=-1$ , but $\measuredangle LA_{1}A=90^{\circ} \Rightarrow \measuredangle IA_{1}L=\measuredangle I_{A}A_{1}L$, which means $AA_1$ is the ex-bisector of $\measuredangle IA_{1}I_{A}$ . From angle chasing $\measuredangle A_{1}AS= \measuredangle SAA_1$ and we have $\measuredangle A'SA=90^{\circ}$ . Now let $H$ is the symmetric point of $A'$ with respect to $S$. From the conditions above it is easy to spot why $H \in AA_1$. But we have, that $HS$ is the perpendicular bisector of line segment $II_A$. So point $H$ is the intersection point of the perpendicular bisector of line segment $II_A$ and the ex-bisector of $\measuredangle IA_{1}I_A$, which means $H \in (IA_{1}I_A)$ $\Rightarrow \measuredangle IA'I_A= \measuredangle I_{A}HI=\measuredangle IA_{1}I_A$ and the conclusion follows.

First note that lines $AA_1$ and $AA'$ are isogonal, therefore point $D$ the symmetric point to point $A_1$ w.r.t. $AI$ lies on $AA'$. We will show that $IDA'I_A$ is cyclic. We have the two following similarities which lead us to the solution $$\triangle ABA_1 \sim \triangle ACA' \Rightarrow AD\cdot AA'=AA_1\cdot AA'=AB\cdot AC$$$$\triangle ABI \sim \triangle ACI_A \Rightarrow AB\cdot AC=AI\cdot AI_A $$Combined together this yields $AD\cdot AA'= AI\cdot AI_A$ and $IDA'I_A$ is cyclic. Using this we conclude: $$\angle IA_1I_A= \angle IDI_A = \angle IA'I_A \enspace _\square$$

Inversion with reflexion on circle with center A and radii $\sqrt{AB \times AC}$. We get that $I$ goes to $I_a$ and $A_1$ goes to $A'$. Reflex $A'$ in $AI$ to $D$ and we obtain that $IDA_1I_a$ is cyclic.