This is Sue Liu's solution proving that the triangle $\mathrm{PQR}$ is equilateral whatever the position of the point $X$. Many congratulations Sue on all your excellent work.

Let $\mathrm{AX}=x$ and $\mathrm{XB}=y$ where we know that $\mathrm{AX}+\mathrm{XB}=\mathrm{AB}$ (constant). Let the points $P,Q$ and $R$ be the centres (centroids) of the triangles $\Delta \mathrm{AXY},\Delta \mathrm{XZB}$ and $\Delta \mathrm{ABC}$ respectively.

We use the fact that the medians of a triangle intersect at the centroid and this point divides the medians in the ratio one third to two thirds.

If we set the point $A$ as the origin, then the points $P$, $Q$ and $R$, being the centroids of the equilateral triangles $\mathrm{AXY}$, $\mathrm{XZB}$ and $\mathrm{ABC}$, have coordinates

${\displaystyle P=(\frac{1}{2}x,\frac{\sqrt{3}}{6}x),}$

${\displaystyle Q=(x+\frac{1}{2}y,\frac{\sqrt{3}}{6}y),\mathrm{and}}$

${\displaystyle R=(\frac{1}{2}(x+y),\frac{-\sqrt{3}}{6}(x+y)).}$

We now show that the lengths $\mathrm{PQ}$, $\mathrm{QR}$ and $\mathrm{RP}$ are equal.

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{{\mathrm{PQ}}^2}& =(x+\frac{1}{2}y-\frac{1}{2}x{)}^2+(\frac{\sqrt{3}}{6}y-\frac{\sqrt{3}}{6}x{)}^2\\ \multicolumn{1}{c}{}& =\frac{x^2+2\mathrm{xy}+y^2}{4}+\frac{y^2-2\mathrm{xy}+x^2}{12}\\ \multicolumn{1}{c}{}& =\frac{x^2+\mathrm{xy}+y^2}{3}.\end{array}}$

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{{\mathrm{QR}}^2}& =(\frac{1}{2}(x+y)-(x+\frac{1}{2}y{)}^2+(-\frac{\sqrt{3}}{6}(x+y)-\frac{\sqrt{3}}{6}y{)}^2\\ \multicolumn{1}{c}{}& =\frac{1}{4}{x}^2+\frac{1}{12}(x^2+4\mathrm{xy}+4y^2)\\ \multicolumn{1}{c}{}& =\frac{x^2+\mathrm{xy}+y^2}{3}.\end{array}}$

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{{\mathrm{RP}}^2}& =(\frac{1}{2}(x+y)-\frac{1}{2}x{)}^2+(-\frac{\sqrt{3}}{6}(x+y)-\frac{\sqrt{3}}{6}x{)}^2\\ \multicolumn{1}{c}{}& =\frac{1}{4}{y}^2+\frac{1}{12}(4x^2+4\mathrm{xy}+y^2)\\ \multicolumn{1}{c}{}& =\frac{x^2+\mathrm{xy}+y^2}{3}.\end{array}}$

As

${\displaystyle \mathrm{PQ}=\mathrm{QR}=\mathrm{RP}=\sqrt{\frac{x^2+\mathrm{xy}+y^2}{3}}}$

for any $x$ it follows that $\Delta \mathrm{PQR}$ is equilateral whatever the position of $X$.