We shall prove that $A_{1}$ is the orthocenter of triangle $I_{a} B C$. Indeed, we have
$$\begin{aligned} \widehat{P E I_{a}} & =180^{\circ}-\widehat{P Q A}-\widehat{E C Q} \\ & =180^{\circ}-\left(90^{\circ}-\frac{1}{2} \widehat{A}\right)-\left(90^{\circ}-\frac{1}{2} \widehat{C}\right) \\ & =\frac{1}{2}(\widehat{A}+\widehat{C})=90^{\circ}-\frac{1}{2} \widehat{B}=\widehat{P B I_{a}} \end{aligned}$$
hence quadrilateral $B E I_{a} P$ is cyclic. Since $\widehat{B P I_{a}}=90^{\circ}$, it follows $B E \perp C I_{a}$. In an analogous way we get $C D \perp B I$, hence $A_{1}$ is the orthocenter of triangle $I_{a} B C$.

It follows that $B A_{1} \parallel C I$, where $I$ is the incenter of triangle $A B C$. Also, $C A_{1} \parallel B I$, hence $B I C A_{1}$ is a parallelogram.
Similarly, $A I B C_{1}$ is a parallelogram, hence $A C_{1} A_{1} C$ is a parallelogram. We obtain that the segments $A A_{1}$ and $C C_{1}$ have the same midpoint. In an analogous way, the segments $B B_{1}$ and $A A_{1}$ have the same midpoint, and the conclusion follows.

Remark. It is clear that $I_{b}, A, I_{c}$ and $I_{c}, B, I_{a}$, and $I_{a}, C, I_{b}$ are collinear.

As in the previous solution, $A_{1}$ is the orthocenter of triangle $I_{a} B C$, hence $I_{a} A_{1} \perp B C$. Similarly, $I_{b} B_{1} \perp C A$ and $I_{c} C_{1} \perp A B$.
The triangles $A B C$ and $I_{a} I_{b} I_{c}$ are orthological, that is the perpendicular lines through $A, B, C$ on $I_{b} I_{c}, I_{c} I_{a}$, and $I_{a} I_{b}$, respectively, are concurrent (as internal bisectors of triangle $A B C$ ). It follows that also, $I_{a} A_{1}, I_{b} B_{1}, I_{c} C_{1}$ are concurrent.