Let $ABC$ be a triangle in which the length of side $AB$ is $4$ units, and that of $BC$ is $2$ units. Let $D$ be the point on $AB$ at distance $3$ units from $A$. Prove that the line perpendicular to $AB$ through $D$, the angle bisector of $\angle ABC$, and the perpendicular bisector of $BC$ all meet at a single point.