Let $ABC$ be an acute triangle, and its incircle touch the sides $AB$ and $AC$ at $K$ and $L$. Let $J$ be the incenter of $\triangle BCD$, where $D$ is a point on $AC$ such that $BD=AB$. Prove that $KL$ bisects $AJ$.
My idea was to draw a line parallel to $KL$ through $J$ intersecting $AC$ at $H$. It is easy to see that $\triangle CHJ \sim \triangle CIB$. I also saw that $\angle JHD=\angle JDH$, implying $JD=JH$. I wanted to prove that $CH=a-c$ using similarity which would easily prove that $AG=JG$.
However, I am not able to finish it. I think the idea of drawing a parallel is a good one, yet I am not able to draw a conclusion? Can anyone help? :)
Hint: introduce point $X$ - reflection of $D$ with respect to $BJ$.