Through the incentre $I$ of triangle $ABC$ a straight line is drawn intersecting $AB$ and $BC$ at points $M$ and $N$, respectively, in such a way that the triangle $BMN$ is acute- angled. On the side $AC$ the points $K$ and $L$ are chosen such that $∠ILA = ∠IMB$ and $∠IKC = ∠INB$. Prove that $AC = AM + KL + CN$.
I have no idea how to start.
I can only see triangle IMC' and ILB' are congruent but not triangle IMC' and IKB'. I wonder if my diagram is different from yours.
Let $P$, $Q$ and $R$ be touching points of the incircle to $BC$, $AC$ and $AB$ respectively.
Also, let $M\in AR$, $N\in PC$, $MR=x$ and $PN=y$.
Thus, since $\Delta MRI\cong\Delta LQI$ and $\Delta NPI\cong\Delta KQI,$ in the standard notation we obtain: $$AM+KL+CN=\frac{b+c-a}{2}-x+x+y+\frac{a+b-c}{2}-y=b=AC$$ and we are done!