Parallel Bisection

In triangle $ABC$ , the perpendiculars from $A$ to the internal bisectors of $\angle CBA$ and $\angle BCA$ meet those bisectors at $X$ and $Y$ . Then which of the following options is correct : $$

A : $XY$ is parallel to $BC$ and $XY\sin B \sin C = BC \sin A$ $$ B : $XY$ is parallel to $BC$ and $BC\sin B \sin C = XY \sin A$ $$ C : $XY$ is parallel to $BC$ and BY = CX $$ D : $XY$ is parallel to $BC$