Brownian motion, why $p\{B_{T_{x+h,x-h}}=x\pm h\}=\frac{1}{2}$?

Question

Let $(B_t)$ a Brownien motion. Let $\tau_a=\inf\{t\geq 0\mid B_t=a\}$ (with $a\neq 0$) and $T_{a,b}=\tau_a\wedge \tau_b$. Suppose $x\in[a,b]$ and $B_0=x$. Let $h>0$ very small. Why do me have $p\{B_{T_{x+h,x-h}}=x\pm h\}=\frac{1}{2}$ ? The argument is the symmetry but I do not understand.