I centered incircle of triangle $ABC$ $(m(\hat{B})=90^\circ)$ touches $\left[AB\right], \left[BC\right], \left[AC\right]$ respectively at $F, D, E$. $\left[CI\right]\cap\left[EF\right]={L}$ and $\left[DL\right]\cap\left[AB\right]=N$. Prove that $\left[AI\right]=\left[ND\right]$.
