<s>[INST] 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?
 
 [/INST] 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
$$\cos\left(30^{\circ}-\frac{\gamma}{6}\right)\sin\frac{\gamma}{3}\sin\frac{\gamma}{6}\sin^2\left(30^{\circ}-\frac{\gamma}{6}\right)\geq0.$$
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