Find the locus of points $P$ such that the distances from $P$ to the sides of a given triangle can themselves form a triangle. 
Find the locus of points $P$ within a given $\triangle ABC$ and such that the distances from $P$ to the sides of the given triangle can themselves be the sides of a certain triangle.

Join $PA,PB,PC$ and let the perpendiculars from $P$ meet $BC,CA,AB$ in $D,E,F$ respectively. Then, $$PA^2 = PE^2 + AE^2 = PF^2 + FA^2$$ $$PB^2 = PD^2 + BD^2 = PF^2 + BF^2$$ $$PC^2 = EC^2 + PE^2 = PD^2 + CD^2$$
$$\therefore PD = \sqrt{PC^2 - CD^2} = \sqrt{PB^2 - BD^2}$$ $$PE = \sqrt{PA^2 - AE^2} = \sqrt{PC^2 - CE^2}$$ $$PF = \sqrt{PB^2 - BF^2} = \sqrt{PA^2 - AF^2} $$
How do I proceed using this ?

HINT:
For points $P$ inside the triangle the condition $d_A=d_B+d_C$ is a part of a line. Can you find two points on it?
So the region should be bounded by three lines. 
