2. $5 P$ is a point inside $\triangle A B C$, from $P$ draw perpendicular segments $P L$, $P M$, $P N$ to the three sides $B C$, $C A$, $A B$, with points $L$, $M$, $N$ being the feet of the perpendiculars. If $P L$, $P M$, $P N$ can form a triangle, find the region where point $P$ is located.
2.5 First, find the locus of points $P$ inside $\triangle ABC$ such that $PL = PM + PN$: Construct the angle bisectors $BB'$ and $CC'$ of $\angle B$ and $\angle C$, respectively. This locus is the line segment $B'C'$. Points $B'$ and $C'$ clearly satisfy the condition $PL = PM + PN$. When point $P$ is on $B'C'$, it is also easy to prove through calculation that it satisfies the condition. When point $P$ is inside $\triangle AB'C'$, we have $PL > PM + PN$.
The angle bisector of $\angle A$ is $AA'$. Therefore, the region of the desired point $P$ is the interior of $\triangle A'B'C'$.