12. In $\triangle \mathrm{ABC}$, a line parallel to $BC$ is drawn through the intersection of the angle bisectors of $\angle \mathrm{B}$ and $\angle \mathrm{C}$. $M$ and $N$ are the points where this line intersects sides $\mathrm{AB}$ and $\mathrm{AC}$, respectively. Try to find the relationship between the segments $\mathrm{MN}, \mathrm{BM}, \mathrm{CN}$. Consider the following cases:
1) Two internal angle bisectors,
2) Two external angle bisectors,
3) One internal angle bisector and one external angle bisector.
Under what conditions do $M$ and $\mathbf{N}$ coincide?
12. 1) Let $BO$ and $CO$ be the angle bisectors of $\triangle ABC$ (Figure 13). It is easy to see that $\triangle BOM$ and $\triangle CON$ are isosceles, hence $\mathrm{MN}=\mathrm{BM}+\mathrm{CN}$.
2) The relation $MN=BM$
$+\mathrm{CN}$ also holds for the external angle bisectors.
3) If one is an internal
angle bisector and the other is an external angle
bisector (Figure 14), then from $\triangle BMO$ and $\triangle CNO$ we can get $MN$ $=\mathrm{CN}-\mathrm{BM}$ (when $\mathrm{CN}>\mathrm{BM}$)
or $MN=BM-CN$
(when $\mathrm{CN}<\mathrm{BM}$). Thus,
in the 3rd case, MN $=|\mathrm{CN}-\mathrm{BM}|$.
Only in the 3rd case
Figure 14

can points $M$ and $N$ coincide, in which case $\triangle ABC$ is isosceles ($AB$ $=AC$).