由旋转得$AB\’=AB$，
$\because \angle BAB\’$的角平分线经过$BC$的中点$E$，
$\therefore \angle EAB\’=\angle EAB$，$CE=BE=\frac{1}{2}BC$，
$\because AB\’=AB$，$AE$平分$\angle BAB\’$，
$\therefore AE$垂直平分$BB\’$，
$\therefore {B\’}E=BE=\frac{1}{2}BC$，
$\therefore \angle EB\’B=\angle {B\’}BC$，
$\because {B\’}C=\frac{1}{2}BC$，
$\therefore {B\’}C={B\’}E=CE$，
$\therefore \triangle {B\’}CE$是等边三角形，
$\therefore \angle {B\’}EC=\angle {B\’}CE=60^{\circ}$，
$\because \angle {B\’}EC=\angle EB\’B+\angle {B\’}BC=2\angle {B\’}BC=60^{\circ}$，
$\therefore \angle {B\’}BC=30^{\circ}$，
$\therefore \angle BB\’C=180^{\circ}-\angle EBB\’-\angle {B\’}CE=90^{\circ}$，$\angle DAB\’=\frac{1}{2}\angle {B\’}BC=15^{\circ}$，
$\because AE\bot BB\’$，${B\’}C\bot BB\’$，
$\therefore AE$∥${B\’}C$，
$\because$四边形$ABCD$是平行四边形，
$\therefore AD$∥$BC$，
$\therefore \angle DAE=\angle AEB=\angle {B\’}CE=60^{\circ}$，
$\therefore \angle EAB\’=\angle DAE-\angle DAB\’=60^{\circ}-15^{\circ}=45^{\circ}$，
$\therefore \angle BAB\’=2\angle EAB\’=90^{\circ}$，
$\therefore BB\’=\sqrt{AB{′}^{2}+A{B}^{2}}=\sqrt{2}AB$，
$\because BB\’=\sqrt{B{C}^{2}-B′{C}^{2}}=\sqrt{B{C}^{2}-(\frac{1}{2}BC)^{2}}=\frac{\sqrt{3}}{2}BC$，
$\therefore \sqrt{2}AB=\frac{\sqrt{3}}{2}BC$，
$\therefore \frac{AB}{BC}=\frac{\sqrt{6}}{4}$，
故答案为：$\frac{\sqrt{6}}{4}$.

由旋转得$AB\’=AB$，
$\because \angle BAB\’$的角平分线经过$BC$的中点$E$，
$\therefore \angle EAB\’=\angle EAB$，$CE=BE=\frac{1}{2}BC$，
$\because AB\’=AB$，$AE$平分$\angle BAB\’$，
$\therefore AE$垂直平分$BB\’$，
$\therefore {B\’}E=BE=\frac{1}{2}BC$，
$\therefore \angle EB\’B=\angle {B\’}BC$，
$\because {B\’}C=\frac{1}{2}BC$，
$\therefore {B\’}C={B\’}E=CE$，
$\therefore \triangle {B\’}CE$是等边三角形，
$\therefore \angle {B\’}EC=\angle {B\’}CE=60^{\circ}$，
$\because \angle {B\’}EC=\angle EB\’B+\angle {B\’}BC=2\angle {B\’}BC=60^{\circ}$，
$\therefore \angle {B\’}BC=30^{\circ}$，
$\therefore \angle BB\’C=180^{\circ}-\angle EBB\’-\angle {B\’}CE=90^{\circ}$，$\angle DAB\’=\frac{1}{2}\angle {B\’}BC=15^{\circ}$，
$\because AE\bot BB\’$，${B\’}C\bot BB\’$，
$\therefore AE$∥${B\’}C$，
$\because$四边形$ABCD$是平行四边形，
$\therefore AD$∥$BC$，
$\therefore \angle DAE=\angle AEB=\angle {B\’}CE=60^{\circ}$，
$\therefore \angle EAB\’=\angle DAE-\angle DAB\’=60^{\circ}-15^{\circ}=45^{\circ}$，
$\therefore \angle BAB\’=2\angle EAB\’=90^{\circ}$，
$\therefore BB\’=\sqrt{AB{′}^{2}+A{B}^{2}}=\sqrt{2}AB$，
$\because BB\’=\sqrt{B{C}^{2}-B′{C}^{2}}=\sqrt{B{C}^{2}-(\frac{1}{2}BC)^{2}}=\frac{\sqrt{3}}{2}BC$，
$\therefore \sqrt{2}AB=\frac{\sqrt{3}}{2}BC$，
$\therefore \frac{AB}{BC}=\frac{\sqrt{6}}{4}$，
故答案为：$\frac{\sqrt{6}}{4}$.