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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.

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  • $\begingroup$ I solved your problem. If you want to see my solution, show us your attempts. $\endgroup$ – Michael Rozenberg Apr 11 at 9:58
  • $\begingroup$ Do you know the AAS (Angle-Angle-Side) Theorem for triangle congruence? $\endgroup$ – Oscar Lanzi Apr 11 at 10:15
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    $\begingroup$ Cleverly done @Michael. Too bad I can't upvote your answer. The OP has to cooperate if your answer is to appear in vote-able form! $\endgroup$ – Oscar Lanzi Apr 11 at 10:24
  • $\begingroup$ @sailormars2016 Please change the title of the question according to your problem. It should be related with the question, not with its context or detailes. $\endgroup$ – Anirban Niloy Apr 11 at 11:03
  • $\begingroup$ @OscarLanzi, how does AAS help? I solved the problem using Ptolemy's theorem. Is there a very simple solution I am missing? $\endgroup$ – Anubhab Ghosal Apr 11 at 11:19
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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!

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