Given the right $\triangle ABC\ :\ A(0,0),B(b,0),C(0,c)$ and $c > b$. Point $D(\frac{bc^{2}}{b^{2}+c^{2}},\frac{b^{2}c}{b^{2}+c^{2}})$. Incenter of $\triangle ABD\ :\ I_{1}(\frac{bc(c+t)}{b^{2}+c^{2}+(b+c)t},\frac{b^{2}c}{b^{2}+c^{2}+(b+c)t})$ with $t=\sqrt{b^{2}+c^{2}}$. Incenter of $\triangle ACD\ :\ I_{2}(\frac{bc^{2}}{b^{2}+c^{2}+(b+c)t},\frac{bc(b+t)}{b^{2}+c^{2}+(b+c)t})$. $d(I_{1},AD)+d(I_{2},AD)=\frac{b^{2}c}{b^{2}+c^{2}+(b+c)t}+\frac{bc^{2}}{b^{2}+c^{2}+(b+c)t}=\frac{1}{4}t$, $3b^{2}-c^{2}=0$ or $\frac{c}{b}=\sqrt{3} \Rightarrow \angle ABC=60^{\circ}$.