Let $R_{(ABY)}$ and $R_{(ACX)}$ denote the circumradius of $(ABY)$ and $(ACX)$, respectively. Note that $\frac{BY}{CX} = \frac{R_{(ABY)}}{R_{(ACX)}}$. By easy angle chase we get that $\triangle BTX \cong \triangle CTY$. And thus, $TC = TX$, which implies that $T$ lies on the $A$ angle bisector (i.e line $I_BI_C$). So by incenter-excenter lemma, we get $\frac{TI_B}{TI_C} = \frac{TB}{TC}$. Also note that $\frac{TB}{\sin{\angle A/2}} = 2R_{(ABY)}$ and $\frac{TC}{\sin{\angle A/2}} = 2R_{(ACX)}$. So dividing, we get $\frac{TI_B}{TI_C} = \frac{TB}{TC} = \frac{R_{(ABY)}}{R_{(ACX)}} = \frac{BY}{CX}$, as desired.