# Problem about incentre of a triangle.

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.

• I solved your problem. If you want to see my solution, show us your attempts. – Michael Rozenberg Apr 11 at 9:58
• Do you know the AAS (Angle-Angle-Side) Theorem for triangle congruence? – Oscar Lanzi Apr 11 at 10:15
• 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! – Oscar Lanzi Apr 11 at 10:24
• @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. – Anirban Niloy Apr 11 at 11:03
• @OscarLanzi, how does AAS help? I solved the problem using Ptolemy's theorem. Is there a very simple solution I am missing? – Anubhab Ghosal Apr 11 at 11:19

## 1 Answer

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!