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From "A Beautiful Journey through Olympiad Geometry":

Part II. Problem 2. Let $I$ be the incenter of $\triangle ABC$. Let $\ell$ be a line [through $I$?] parallel to $\overline{AB}$, that intersects the sides $\overline{CA}$ and $\overline{CB}$ at $M$ and $N$, respectively. Prove that $$|\overline{AM}| + |\overline{BN}| = |\overline{MN}|$$

I think this problem is wrong because I could prove that $|\overline{MN}|$ is not equal to $|\overline{AM}|+|\overline{BN}|$. And what is the role of incenter here?

Edited by @Blue. Note: The source does not mention that $\ell$ passes through $I$. This appears to be a simple omission, as the stated relation is readily shown to hold only with that assumption.

Blue
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user459947
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    Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – gebruiker Jul 04 '17 at 07:31
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    Also, on this site we use MathJaX to format our maths. Here you can find a basic tutorial. – gebruiker Jul 04 '17 at 07:32
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    I assume that the line $\ell$ should go through $I$. Also, could you maybe not use a capital $I$ and a small $l$, they are confused way to easily. If you have a proof that this is wrong, why not give it, then we can see what you are talking about. – Dirk Jul 04 '17 at 07:33

2 Answers2

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Let's change the notation somewhat so that the symmetry of the figure is more apparent. Specifically, draw lines $A_c B_c$, $B_a C_a$, $C_b A_b$ parallel to $AB$, $BC$, $CA$ respectively, and such that all pass through the incenter $I$. Then this creates three similar triangles: $$\triangle A_c I C_a \sim \triangle A_b B_a I \sim \triangle I B_c C_b \sim \triangle ABC.$$ Consequently, the resulting figure also contains three parallelograms, $$AA_bIA_c, \quad BB_cIB_a, \quad CC_aIC_b.$$ But since the distance between each pair of parallel lines is the same and is the inradius $r$, the parallelograms are actually rhombi, and in particular $$AA_c = A_cI, \quad BB_c = B_cI,$$ thus $$A_cB_c = AA_c + BB_c,$$ which proves the desired relationship.

heropup
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I'll prove the following:

Given that $\overline{MN}\parallel\overline{AB}$, then $$a+b=m+n \quad\iff\quad \text{$P$ is the incenter of $\triangle ABC$}$$

enter image description here

The parallelism condition tells us right away that $$\angle 1 \cong \angle 2 \qquad\text{and}\qquad \angle 3 \cong \angle 4$$

So ...

$$\begin{align} a+b = m+n \quad&\iff\quad \text{($\Rightarrow$, we may assume; $\Leftarrow$ from below, we deduce) $a=m$ and $b=n$.} \\ \quad&\iff\quad \text{$\triangle AMP$ and $\triangle BNP$ are isosceles} \\ \quad&\iff\quad \angle \theta \cong \angle 1\cong\angle 2 \quad\text{and}\quad \angle \phi \cong \angle 3\cong\angle 4 \\ \quad&\iff\quad \text{$\overline{AP}$ and $\overline{BP}$ are angle bisectors} \\ \quad&\iff\quad \text{$P$ is the incenter of $\triangle ABC$} \end{align}$$

Blue
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