Relation involving lengths determined by a line (through the incenter?) of a triangle 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. 
 A: 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.
A: I'll prove the following:

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


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