Problem
Let $I$ be the incenter of $\Delta ABC$. Produce $AI$, $BI$ and $CI$ to $J_A$, $J_B$ and $J_C$ respectively such that $J_BAJ_C$, $J_CBJ_A$ and $J_ACJ_B$ are straight lines. Prove that $J_A$, $J_B$ and $J_C$ are the excenters of $\Delta ABC$.
Solution by using barycentric coordinates
Let $|BC|=a$, $|CA|=b$ and $|AB|=c$. Recall that in barycentric coordinates (not necessarily normalised), $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$ and $I=(a,b,c)$.
Since $AIJ_A$, $BIJ_B$ and $CIJ_C$ are straight lines, we have $J_A=(r_a,b,c)$, $J_B=(a,r_b,c)$ and $J_C=(a,b,r_c)$, where $r_a$, $r_b$ and $r_c$ are parameters to be determined.
Since $J_BAJ_C$, $J_CBJ_A$ and $J_ACJ_B$ are straight lines, we have $$ 0= \begin{vmatrix} 1 & 0 & 0 \\ a & b & r_c \\ a & r_b & c \end{vmatrix} =bc-r_cr_b \implies r_br_c=bc\\ 0= \begin{vmatrix} 0 & 1 & 0 \\ r_a & b & c \\ a & b & r_c \end{vmatrix} =ca-r_ar_c \implies r_cr_a=ac\\ 0= \begin{vmatrix} 0 & 0 & 1 \\ a & r_b & c \\ r_a & b & c \end{vmatrix} =ab-r_br_a \implies r_ar_b=ab $$
Solving, we have $(r_a,r_b,r_c)=\pm(a,b,c)$. Note that $(r_a,r_b,r_c)=+(a,b,c)$ implies $J_A=J_B=J_C=I$. We should take $(r_a,r_b,r_c)=-(a,b,c)$. Then we have $J_A=(-a,b,c)$, $J_B=(a,-b,c)$ and $J_C=(a,b,-c)$, which is exactly the barycentric coordinates of the excenters.
Question
I can finish the proof by using barycentric coordinates as given above. I would like to know if there is an elementary proof.