SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Let $P$ be the foot of perpendicular from $I_a$ to $BC$, $D$ the midpoint of $BC$ and $S_a$ the reflection of $A$ with respect to $S$. The line through $I$ parallel to $BC$ cuts $AP$, $A{A}^{\prime }$, $AV$ at $Q$, $R$, $L$, respectively. Redefine $A^{\prime }$ as follow: Let $I$ be the incenter and $S$ the Spieker point of triangle $ABC$. Assume that $A^{\prime }\equiv AS\cap BC$ 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 $P{S}_a\parallel RV$ and $P{S}_a\perp BC$ and since $DI\parallel AP$, $PQD{S}_a$ is a parallelogram. This implies that $DQ\parallel P{S}_a$ and $DQ\perp BC$. As a result $Q\in DT$ and $VR$ is perpendicular bisector of $LQ$ then $$V(L,Q,R,P)=-1$$ If $RV$ cuts $A{I}_a$, $AP$ 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 $$X{A}^{\prime }\parallel I_{a}P\parallel X_{a}V\text{ and }X{A}^{\prime }\perp BC\text{.}$$ 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 $△ABC$.