Recall that the incenter of $\,$$\Delta$$ABC$ $\,$is the center of its incircle and is located at the intersection of the angle bisectors of $\,$$\Delta$$ABC$$\,$.
Morley showed that intersecting adjacent trisectors of $\,$$\Delta$$ABC$$\,$ yields an equilateral triangle (the Morley triangle) inside $\,$$\Delta$$ABC$
$\,$. $\;$Define the Morley center of $\,$$\Delta$$ABC$$\,$$\,$to be the center of its Morley triangle.
The definitions above as well as a little experimentation suggest that both of the following should be true. Are there simple proofs for either?$\;$ Thanks.
Questions: $\,$1) Does the Morley center lie inside the incircle?
$\qquad$$\qquad$ $\,$ 2) Does the incenter lie inside the Morley triangle?
The vertices of the Morley's equilateral triangle have trilinear coordinates given by $$ M_A = \left[1,2\cos\frac{C}{3},2\cos\frac{B}{3}\right],\quad M_B=\left[2\cos\frac{C}{3},1,2\cos\frac{A}{3}\right],\quad M_C=\left[2\cos\frac{B}{3},2\cos\frac{A}{3},1\right]$$ where $2\cos\frac{x}{3}\in (1,2)$ for any $x\in\{A,B,C\}$. Since the trilinear coordinates of the incenter are given by $I=[1,1,1]$, it follows that the incenter always lies inside $M_A,M_B,M_C$.
The trilinear coordinates of the Morley centre $X(356)$ are given by: $$ X(356)=\left[\cos\frac{A}{3}+2\cos\frac{B}{3}\cos\frac{C}{3},\ldots,\ldots\right]$$ but these numbers always lie in the interval $\left(2,1+\cos\frac{\pi}{9}+\cos\frac{2\pi}{9}=2.705737\ldots\right)$, hence the Morley centre always lies inside the incircle, too.