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 ?
 A: The essential quantities are the distances $d_a$, $d_b$, $d_c$ of $P$ from the sides $a$, $b$, $c$ of the given triangle $\triangle$. We therefore should describe the problem in such a format that these quantities become simple, preferably linear, functions of the data.
Let the incircle $\kappa$ of $\triangle$ have radius $r$ and its center  at the origin $O$. Let 
$$r(\cos\alpha,\sin\alpha),\quad r(\cos\beta,\sin\beta),\quad r(\cos\gamma,\sin\gamma)$$
be the  points of contact of $\kappa$ with the sides $a$, $b$, $c$ of $\triangle$. The distance $d_a$ of $P=(x,y)\in\triangle$ from the  side $a$ is then given by
$$d_a=r-(x\cos\alpha+y\sin\alpha)\ ,$$
and similarly for $d_b$ and $d_c$. The triangle inequality $d_a\leq d_b+d_c$ then leads to the condition
$$x(\cos\beta+\cos\gamma-\cos\alpha)+y(\sin\beta+\sin\gamma-\sin\alpha)\leq r\ .$$
The boundary of the set of points fulfilling this condition is the  line $$\ell_a:\qquad x(\cos\beta+\cos\gamma-\cos\alpha)+y(\sin\beta+\sin\gamma-\sin\alpha)= r\ ,\tag{1}$$ and we obtain two more such lines $\ell_b$, $\ell_c$. It follows that the feasible region (the "locus" in question) is a triangle $T$ containing obviously the point $O$. We now have to find the vertices $L_\iota$ of $T$. To this end I supplement $(1)$  with the equation for $\ell_b$:
$$x(\cos\gamma+\cos\alpha-\cos\beta)+y(\sin\gamma+\sin\alpha-\sin\beta)= r\ .\tag{2}$$
Adding $(1)$ and $(2)$ gives $x\cos\gamma+y\sin\gamma=r$. This says that the vertex $L_c:=\ell_a\wedge\ell_b$ is lying on $c$. Subtracting $(1)$ from $(2)$ leads to
$$-x\sin{\alpha+\beta\over2}+y\cos{\alpha+\beta\over2}=0\ ,$$
which says that $L_c$ is lying on the angle bisector $C\vee O$ of $\triangle$.
I can leave the rest to you.
A: If the triangle is equilateral, then the sum of the distances from an arbitrary point inside the triangle to the three sides is equal to the height of the triangle, which is a constant. So the three perpendiculars are the lengths of a triangle when all of them are not longer than half the height of the given triangle. The point $P$ must lie inside the midpoint triangle of the given triangle.
For a triangle in general, I used Excel to generate random points and plot all points satisfying the condition.


My conclusion is that $P$ must be a point inside the incentral triangle of the given triangle.
We first consider the limiting case when the sum of the lengths of two perpendiculars is equal to that of the third one.

Let $BE$ be the bisector of $\angle ABC$ and $CF$ the bisector of $\angle ACB$, where $E$ lies on $AC$ and $F$ lies on $AB$. Let $G$ be a point on $EF$.
Let $X_1$, $X_2$ and $X_3$ be the feet of perpendicular to $BC$ from $E$, $F$ and $G$ respectively. Let $Y_2$ and $Y_3$ be the feet of perpendicular to $AC$ from $F$ and $G$ respectively. Let $Z_1$ and $Z_3$ be the feet of perpendicular to $AB$ from $E$ and $G$ respectively.
Note that $EX_1=EZ_1$ and $FX_2=FY_2$.
We have $\displaystyle \frac{GX_3-EX_1}{FX_2-EX_1}=\frac{EG}{EF}$, $\displaystyle \frac{EZ_1-GZ_3}{EZ_1}=\frac{EG}{EF}$ and $\displaystyle \frac{GY_3}{FY_2}=\frac{EG}{EF}$.
Therefore,
$$\frac{GX_3-EX_1}{FX_2-EX_1}=\frac{EZ_1-GZ_3}{EZ_1}=\frac{GY_3}{FY_2}$$
$$\frac{GX_3-EX_1}{FX_2-EX_1}=\frac{EX_1-GZ_3}{EX_1}=\frac{GY_3}{FX_2}$$
So,
$$GX_3-EX_1=GY_3-(EX_1-GZ_3)$$
$$GX_3=GY_3+GZ_3$$
(Note: this equality holds also for external points of division of $EF$ if we consider negative lengths.)
If $P$ is a point inside $BCEF$, let $Q$ be a point on $EF$ such that $QP$ is perpendicular to $BC$. The distance from $P$ to $BC$ is less than the distance from $Q$ to $BC$. The distance from $P$ to $AC$ is greater than the distance from $Q$ to $AC$ and the distance from $P$ to $AB$ is greater than the distance from $Q$ to $AB$. We can now conclude that the distance from $P$ to $BC$ is less than the sum of the distance from $P$ to $AB$ and the distance from $P$ to $AC$.
Let $D$ be a point on $BC$ such that $AD$ bisects $\angle BAC$.

Repeating the above arguments, the perpendiculars from $P$ to the three sides of the triangle are the lengths of a triangle if and only if $P$ lies inside $\triangle DEF$.
A: 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. 
A: Let PA’, PB’ and PC’ the perpendicular distances to the sides BC, CA and AB. WOLG, we can assume PA’ is the longest and PB’ the shortest among the three. For this to happen, P should be inside the region bounded by $\alpha$ (the angle bisector of $\angle BAC$), $\beta$ (the angle bisector of $\angle ABC$) and the line AC.

(1) Form the //gm PA’QB’; (2) Let the red circle (center = A’, radius = A’Q) cut AP at Q’; (3) Let the blue circle (center = P, radius = PC’) cut PA’ at D. 
According to triangular inequality, we need $PC’ (= PD)$ $\gt PQ’$. The limiting case is when $PC’ = PQ’$ (i.e. when D and Q’ are coincide). For this reason, we draw the line $\lambda$, which is a line parallel to BC and at a distance of PA’ = PQ’ + Q’A. 
The locus of P is therefore the purple region bounded by $\alpha$, $\beta$ and $\lambda$.
