Saudi Arabian Mathematical Competitions

Let $P$ be the foot of perpendicular from $I_{a}$ to $B C$, $D$ the midpoint of $B C$ and $S_{a}$ the reflection of $A$ with respect to $S$. The line through $I$ parallel to $B C$ cuts $A P$, $A A^{\prime}$, $A V$ at $Q$, $R$, $L$, respectively.
Redefine $A^{\prime}$ as follow: Let $I$ be the incenter and $S$ the Spieker point of triangle $A B C$. Assume that $A^{\prime} \equiv A S \cap B C$ and $V$ is the reflection of $P$ with respect to $A^{\prime}$. Then we need to prove that $A^{\prime}$ is also satisfy the given construction.

Since $D$ is clearly the midpoint of $I S_{a}$, then $A^{\prime}$ is the midpoint of $R S_{a}$, so $P S_{a} \parallel R V$ and $P S_{a} \perp B C$ and since $D I \parallel A P$, $P Q D S_{a}$ is a parallelogram. This implies that $D Q \parallel P S_{a}$ and $D Q \perp B C$.
As a result $Q \in D T$ and $V R$ is perpendicular bisector of $L Q$ then
$$V(L, Q, R, P) = -1$$
If $R V$ cuts $A I_{a}$, $A P$ at $X_{a}$, $P_{a}$, we also get
$$V(L, Q, R, P) = (A, Q, P_{a}, P) = (A, T, X_{a}, I_{a}) = -1$$
Then $X$ is midpoint of $I_{a} X_{a}$ which implies that
$$X A^{\prime} \parallel I_{a} P \parallel X_{a} V \text{ and } X A^{\prime} \perp B C \text{.}$$
It's similar to $B^{\prime}$ and $C^{\prime}$. So $A A^{\prime}$, $B C^{\prime}$, $C C^{\prime}$ are concurrent at the Spieker point $S$ of $\triangle A B C$.