Suppose that the perpendicular bisectors meet at M. WLOG Suppose $A_1M\geq A_2M \geq A_3M$. Suppose $A_1M>A_2M$. Then $A_1M>A_2M\geq A_3M$. For all $X\in (A_2A_3)$ we have $MX \leq A_2M$. But M lies on the perpendicular bisector of $A_1B_1$ and thus $MB_1=MA_1$. $MA_1=MB_1\leq MA_2<MA_1$ is a contradiction. So $A_1M=A_2M$. Suppose $A_2M>A_3M$. We have $MA_2=MB_2$. Because $MA_1>MA_3$, on $(A_1A_3)$ there is no point X such as $MX=MA_1$. This leads to a contradiction. So $MA_1=MA_2=MA_3$. On $(A_2A_3)$ there is no point X such as $MX=MA_1$. Again a contradiction. If you can find your way through all the assumptions you can see that now the problem is solved.

I also suppose that the perpendicular bisectors of these segments are concurrent in a point $ O $. we have that in a triangle $ ABC $ with $ X\in (BC) $ the inequality $ AX<max(AB,AC) $. we see that using this we obtain a contradiction immediately.