Let $X', Y' \in AD$ such that $\angle{KXX'} = \angle{KYY'} = 90^{\circ}$; then it suffices to show that $X' = Y'$. Note that $\triangle{BIK}$ is similar to the triangle formed by $BI, BA, XK$, so $\frac{BX}{BI} = \cos \angle{B} = \frac{AX'}{AI_B}$, where $I_B \in AD$ such that $II_B \parallel AB$. By LoS, $AI_B = BI \cdot \frac{\sin \frac{\angle{B}}{2}}{\cos \angle{B}} \implies AX' = BI \sin \frac{B}{2}$. By LoS on $\triangle{IBC}$, this equals $CI \sin \frac{\angle{C}}{2} \implies AX' = AY'$ as desired. $\square$