Distances from Morley triangle to edges of the original triangle Morley's trisector theorem states that in any triangle, the three points of intersection of the adjacent angle trisectors form an equilateral triangle, as illustrated in the left-hand diagram below.  Let's call this equilateral triangle the Morley triangle and its edge length $m$. 
I am interested the distances from the Morley triangle to the edges of the original triangle as illustrated in blue in the right-hand diagram.
Empirically it seems that these distances lie between $\frac{\sqrt 3}{2}m$ and $m$, depending on the angles of the original triangle. Is there simple proof?
 
 A: Let $a,b,c$ be the sides of the triangle and, for simplicity, $A=3\alpha, B=3\beta, C=3\gamma$ be the angles, and hence $\alpha+\beta+\gamma=60^\circ$.

Regarding the triangle $\triangle AEC$ we have (recall the law of sines) $$\frac{AE}{\sin\gamma}=\frac{b}{\sin(180^\circ-\alpha-\gamma)}\iff AE=\frac{b\cdot \sin \gamma}{\sin(\alpha+\gamma)}$$ Consider now the right triangle $\triangle EHA$, where $$\sin\alpha=\frac{EH}{AE}\iff EH=AE\cdot \sin\alpha=\frac{b\cdot\sin\alpha\cdot\sin\gamma}{\sin(\alpha+\gamma)}$$ Now, have a looke here, where they prove (using the same notation) that $m:=EF=8R\sin\alpha\sin\beta\sin\gamma$. Thus $$\frac{m}{EH}=\frac{8R\sin\beta\cdot\sin(\alpha+\gamma)}{b}=\frac{8R\sin\beta\cdot \sin(60^\circ-\beta)}{b}=\frac{8R\cdot\sin\beta\cdot\cos(\beta+30^\circ)}{b}$$
The expansion of $\cos(x+y)=\cos x\cos y-\sin x\sin y$, yields $$\frac{m}{EH}=\frac{8R\sin\beta\cdot\left(\frac{\sqrt{3}}2\cos \beta-\frac12\sin\beta\right)}{b}=\frac{2R\cdot\left(\sqrt{3}\sin(2\beta)-2\sin^2\beta\right)}b$$ Finally, in virtue of the law of Sines, $b=2R\cdot \sin(3\beta)$. Thus $$\frac{m}{EH}=\frac{\sqrt{3}\sin(2\beta)-2\sin^2\beta}{\sin(3\beta)}:=f(\beta)$$ Considering that $60^\circ\geqslant\beta\geqslant0^\circ$, you can prove your inequality using calculus.

EDIT: The acute observation @John Bentin made yields $$f(\beta)=\frac1{\cos\left(\beta-30^\circ\right)}$$
Since we are dealing with the cosine function, we know $\cos(\beta-30^\circ)$ will atain its maximum at $\beta=30^\circ$ which, thus is the minimum of $f$. Furthermore, since $0\leqslant\beta\leqslant 60^\circ$, the cosine function will atain ist minimum at the borders, i.e. at $\beta=0^\circ$ or $\beta=60^\circ$. This leads to the conclusion , that the minima and maxima are resprecitvely $\displaystyle \frac1{\cos 0^\circ}=1$ and $\displaystyle \frac1{\cos30^\circ}=\frac2{\sqrt{3}}$.
A: Let $\Delta XYZ$ ($X$ is opposite to $A$, $Y$ is opposite to $B$ and $Z$ is opposite to $C$)  be the Morley's triangle of $\Delta ABC.$
Thus, $$S_{\Delta ZAB}=\frac{c^2\sin\frac{\alpha}{3}\sin\frac{\beta}{3}}{2\sin\frac{\alpha+\beta}{3}}=\frac{ch_Z}{2},$$
which gives $$h_Z=\frac{c\sin\frac{\alpha}{3}\sin\frac{\beta}{3}}{\sin\frac{\alpha+\beta}{3}}=\frac{2R\sin\gamma\sin\frac{\alpha}{3}\sin\frac{\beta}{3}}{\sin\left(60^{\circ}-\frac{\gamma}{3}\right)}.$$
We need to prove that $$\frac{\sqrt3}{2}m\leq h_Z\leq m.$$
The right inequality.
We need to prove that:
$$\frac{2R\sin\gamma\sin\frac{\alpha}{3}\sin\frac{\beta}{3}}{\sin\left(60^{\circ}-\frac{\gamma}{3}\right)}\leq8R\sin\frac{\alpha}{3}\sin\frac{\beta}{3}\sin\frac{\gamma}{3}$$ or
$$\sin\gamma\leq4\sin\frac{\gamma}{3}\sin\left(60^{\circ}-\frac{\gamma}{3}\right),$$ which is true because
$$4\sin\frac{\gamma}{3}\sin\left(60^{\circ}-\frac{\gamma}{3}\right)-\sin\gamma=8\sin\frac{\gamma}{3}\sin^2\left(15^{\circ}-\frac{\gamma}{6}\right)\cos\left(30^{\circ}+\frac{\gamma}{3}\right)\geq0.$$
The left inequality.
We need to prove that:
$$4\sqrt3R\sin\frac{\alpha}{3}\sin\frac{\beta}{3}\sin\frac{\gamma}{3}\leq\frac{2R\sin\gamma\sin\frac{\alpha}{3}\sin\frac{\beta}{3}}{\sin\left(60^{\circ}-\frac{\gamma}{3}\right)}$$ or
$$2\sqrt3\sin\frac{\gamma}{3}\sin\left(60^{\circ}-\frac{\gamma}{3}\right)\leq\sin\gamma,$$ which is equivalent to
$$\cos\left(30^{\circ}-\frac{\gamma}{6}\right)\sin\frac{\gamma}{3}\sin\frac{\gamma}{6}\sin^2\left(30^{\circ}-\frac{\gamma}{6}\right)\geq0.$$
A: The result drops out as a corollary of the following constructive proof of Morley’s theorem.
Let $\alpha$, $\beta$, and $\gamma$ be positive angles summing to $\frac13\pi$. We start with a regular triangle $A_0B_0C_0$ of side $m$ and attach to it projecting whiskers $A_0A_1$, $B_0B_1$, and $C_0C_1$ of respective lengths $m\sin(\alpha+\frac13\pi)$, $m\sin(\beta+\frac13\pi)$, and $m\sin(\gamma+\frac13\pi)$, making angles with the sides of the triangle as follows:
$$(\text{at }A_0):\qquad\angle B_0A_0A_1=\beta+\tfrac56\pi-\gamma,\qquad\angle C_0A_0A_1=\gamma+\tfrac56\pi-\beta;$$
$$(\text{at }B_0):\qquad\angle C_0B_0B_1=\gamma+\tfrac56\pi-\alpha,\qquad\angle A_0B_0B_1=\alpha+\tfrac56\pi-\gamma;$$
$$(\text{at }C_0):\qquad\angle A_0C_0C_1=\alpha+\tfrac56\pi-\beta,\qquad\angle B_0C_0C_1=\beta+\tfrac56\pi-\alpha.$$
(As required, each of these angle pairs add to $\frac53\pi$, the external angle at each vertex of $\triangle A_0B_0C_0$.) Next, lines perpendicular to $A_0A_1$, $B_0B_1$, and $C_0C_1$ are drawn through $A_1$, $B_1$, and $C_1$ respectively, meeting to form the sides of a triangle $ABC$, where $A_1$ lies on $BC$, $B_1$ lies on $CA$, and $C_1$ lies on $AB$.
The figure thus formed comprises three pentagons $AB_1B_0C_0C_1$, $BC_1C_0A_0A_1$, and $CA_1A_0B_0B_1$, of whose fifteen sides twelve are distinct  and three ($A_0A_1$, $B_0B_1$, $C_0C_1$) are shared; three sides ($B_0C_0$, $C_0A_0$, $A_0B_0$)  form the original regular triangle, and six sides comprise three contiguous collinear pairs ($BA_1$ with $A_1C$, $CB_1$ with $B_1A$, and $AC_1$ with $C_1B$).
Each pentagon is determined by the angles at four consecutive vertices together with the distances between them. In particular, the pentagon $AB_1B_0C_0C_1$ is fixed by $$\angle B_1B_0C_0=\gamma+\tfrac56\pi-\alpha,\qquad\angle B_0C_0C_1=\beta+\tfrac56\pi-\alpha,$$ and the right angles at $B_1$ and $C_1$ along with the distances $|B_1B_0|=m\sin(\beta+\frac13\pi)$, $|B_0C_0|=m$, and $|C_0C_1|=m\sin(\gamma+\frac13\pi)$.
This pentagon may also be constructed from scratch in the following way (where we presumptuously use the same lettering, for ease of comparison). First, on a base $B_0C_0$ (of length $m$), draw the sides $AB_0$ and $AC_0$ of a triangle to make respective base angles $\gamma+\frac56\pi$ at $B_0$ and $\beta+\frac56\pi$ at $C_0$, so that the apex angle at $A$ is $\alpha$. By the sine rule,$$|AB_0|=m\sin(\beta+\tfrac13\pi)\operatorname{cosec}\alpha\quad\text{and}\quad|AC_0|=m\sin(\gamma+\tfrac13\pi)\operatorname{cosec}\alpha.$$Next, externally adjoin to triangle $AB_0C_0$ right-angled triangles $AB_0B_1$ and $AC_0C_1$ with hypotenuses $AB_0$ and $AC_0$, such that
$$|B_0B_1|=m\sin(\beta+\tfrac13\pi)\quad\text{and}\quad |C_0C_1|=m\sin(\gamma+\tfrac13\pi),$$ with right angles at $B_1$ and $C_1$. It follows from the sine ratios $|B_0B_1|/|AB_0|$ and $|C_0C_1|/|AC_0|$ that $\angle B_0AB_1=\angle C_0AC_1=\alpha$.
It is easy to check that the new pentagon $AB_1B_0C_0C_1$ is the same as the original one of the same name. Hence, in the original pentagon, the diagonals $AB_0$ and $AC_0$ trisect the angle at the vertex $A$. Similar results apply to the pentagons $BC_1C_0A_0A_1$ and $CA_1A_0B_0B_1$, yielding the Morley picture.
To answer the question now, observe that (for example) the ratio $|A_0A_1|/|A_0B_0|=\sin(\alpha+\frac13\pi)$ attains its maximum value $1$ when $\alpha=\frac16\pi$, and tends to its infinum $\frac12\surd3$ as $\alpha$ approaches its bounds $0$ or $\frac13\pi$.
