(Inspired by this problem)
In $\triangle ABC$ let the incircle touch $BC$, $CA$ and $AB$ at $D$, $E$ and $F$ respectively. Let line $AD$ cut an incircle at point $L$ and line $LB$ and $LC$ cut incircle at points $M$ and $N$ respectively. Prove that lines $ME$, $NF$ and $ALD$ concur.
I was able to prove this brute-force by using trigonometric form of Ceva's theorem applied to triangle $\triangle LMN$ with lines $ME$, $NF$ and $LD$ as cevians. It's not too complex (about one page) and the calculation of three different sine ratios can be easily followed.
However, it leaves the impression that the solution could be simpler. Is there a shorter way to prove the statement without trigonometry?