Let $ABC$ be a triangle with $I_A$, $I_B$, and $I_C$ as excenters. Prove that triangle $I_AI_BI_C$ has orthocenter $I$ and that triangle $ABC$ is its orthic triangle.
I found this question in $\underline{\text{Euclidean Geometry in Mathematical Olympiads}}$ by $\underline{\text{Evan Chen}}$. It is an exercise question. I have tried my solution and wanted to verify it.
Here is what I have tried:
My Approach:
Let the ray $AI$ meet the circumcircle of $ABC$ at $L$ and let $I_A$ be the refection of $I$ with respect to $L$.
Using the fact that the $A-$excentre is the reflection of the incentre with respect to the point where the ray joining $A$ and incentre meets the circumcircle of $ABC$, it can be asserted that $I_A$ is the $A-$excentre for triangle $ABC$.
On drawing the $A-$excircle, it can be seen that $IBI_AC$ is cyclic with $II_A$ as diameter, thus $\angle IBI_A=90^\circ$.
Doing similar construction for ray $CI$ and drawing the $C-$excircle gives $IAI_CB$ to be cyclic with $II_C$ as diameter, thus $\angle IBI_C=90^\circ$
$\angle IBI_A+\angle IBI_C=180^\circ\Rightarrow I_C,B,I_A$ are collinear and $IB\perp I_CI_A$.
Since the $B-$excentre $(I_B)$ would be the reflection of $I$ about the point where ray $BI$ will meet circumcircle of $ABC$, $I_B,I,B$ would be collinear. Thus $I_BB$ is altitude from $I_B$ to $I_CI_A$.
Similarily $I_CC$ is altitude from $I_C$ to $I_AI_B$ and $I_AA$ is altitude from $I_A$ to $I_BI_C$ all meeting at I, therefore $I$ is the orthocentre for triangle $I_AI_BI_C$ with $ABC$ as its orthic triangle.
Please check if my solution is missing something or if something needs to be added to it. Also feel free to provide any alternative solutions. Your help would be highly appreciated.
THANKS