Given $\triangle ABC\ :\ A(0,a),B(-b,0),C(c,0)$. Then $A_{1}(\frac{c-2b}{3},0),A_{2}(\frac{2c-b}{3},0)$ and $B_{1}(\frac{2c}{3},\frac{a}{3}),B_{2}(\frac{c}{3},\frac{2a}{3})$ and $C_{1}(-\frac{b}{3},\frac{2a}{3}),C_{2}(-\frac{2b}{3},\frac{a}{3})$. The perpendicular bisector of $A_{1}A_{2}\ :\ x=\frac{c-b}{2}$ must be the same as the perpendicular bisector of $B_{2}C_{1}\ :\ x=\frac{c-b}{6}$, so $x=0$ and $b=c$. The perpendicular bisector of $C_{1}C_{2}\ :\ y-\frac{a}{2}=-\frac{b}{a}(x+\frac{b}{2})$ cuts $x=0$ in the point $(0,\frac{a^{2}-b^{2}}{2a})$. The perpendicular bisector of $A_{1}C_{2}\ :\ y-\frac{a}{6}=\frac{c}{a}(x-\frac{c-4b}{6})$ cuts $x=0$ in the point $(0,\frac{a^{2}+3b^{2}}{6a})$. Both points on the y-axis must be the same, so $a=\sqrt{2}c$. Then $\triangle ABC$ is equilateral.