at a first glance, an easy angle-chasing shows that $\angle B=60$ and that $CF$ must be altitude. now notice that $BE$, intersects the circumcircle of triangle at $T$ (midpoint of arc $AC$) and that $T$ is reflection of $O$ wrt $BC$. so $TOXA$ is cyclic.....

In no case, $OYZ$ and $OZX$ are equilateral.

Since $\angle XDB=90$ and angle DXB= 60, then angle XBD=30. This means that angle ABX=30. Furthermore, since angle BXA= 120, then angle BAX=30 and triangle AXB is isosceles. Similarly, since angle CYB=120, then angle BCY=30 and triangle CYB is isosceles. Notice that CF is perpendicular to AB because angle FCB=30 and angle FBC=60. Now form an equilateral triangle PQR where X, Y, and Z are the midpoints of sides PQ, QR, and RP respectively. Notice that now we have to prove that P, Q,or R is the circumcenter of triangle ABC. Extend the sides of the triangle PQR until they intersects the sides of triangle ABC. (The extension of PQ will intersect AB at F* and AC at F**. The extension of QR intersects CB at D* and AB at D**. The extension of PR intersects AC at E* and CB at E**.) Since CF is parallel to F**F*, then F**F* is perpendicular to AB. Similarly, D**D* is perpendicular to side CB, but E*E** is not perpendicular to side AC because triangle ABC is scalene. Since Q is the only vertex from triangle PQR that forms by the intersection of two perpendicular lines of triangle ABC, then we will focus on proving that Q is always the circumcenter. Since XAB is isosceles and XF* is perpendicular to side AB, then F* must be the midpoint of side AB. Also, since angle BYD**= 120 and angle D**BY=30, then triangle BYD** is isosceles. We know that CY=BY and now we learned that BY=YD**. Hence, by forming line CD** we get that triangle CYD is isosceles with angle CD**Y=30. Since D**D* is the angle bisector of angle CD**B, triangle CD**B is isosceles, and D**D* is perpendicular to side CB, then D* is the midpoint of CB. Notice that we have proved that F**F* and D**D* are the perpendicular bisectors of AB and CB respectively. We know that the intersection of F**F* and D**D* is the circumcenter of triangle ABC. Since the intersection of these two segments is Q, then we have finished the proof. Is this proof correct?

If $\triangle XYZ$ is equilateral, then $\angle ZXY = 60º$, and $\angle BXD = 90º - \angle \frac{ABC}{2}= \angle ZXY = 60º$, so $\angle \frac{ABC}{2} = 30º$, and $\angle ABC = 60º$, implying, $\angle ABE = \angle CBE = 30º.$ Now observe $\triangle BFY:$ $\angle BFY + \angle BYF + \angle FBY = 180º$, so, $\angle BFY + 60º + 30º = 180º \implies \angle BFY = \angle BFC = \angle AFC= 90º.$ Furthermore, $\angle BAX = \angle XBA = 30º$, and $\angle YBC = \angle YCB = 30º$, then $X$ is at $AB$ perpendicular bissector, and $Y$ is at $BC$ angle bissector, therefore, $OX \perp AB$ and $CF \perp AB$. Thus $OX||ZY$. We have too that $OY$ and $AD$ are perpendicular to $BC$ from this we conclude that $OY||ZX$, so $OYZX$ is paralelogram, then $OY = ZX$ and $OX = ZY$. Thus, triangle $OXY$ is congruent to triangle $XYZ$, from the $Side$ $Side$ $Side$ case, and $OXY$ is equilateral. Q.E.D.