$\because \angle BAC=90^{\circ}$，$AB=AC$，$AD\bot BC$于点$D$，
$\therefore BD=AD$，$\angle B=\angle C=\angle BAD=\angle DAC=45^{\circ}$，
$\because AD\bot BC$于点$D$，$\angle EDF=90^{\circ}$，
$\therefore \angle BDE+\angle ADE=\angle ADE+\angle ADF=90^{\circ}$，
$\therefore \angle BDE=\angle ADF$，
在$\triangle BED$与$\triangle AFD$中，
$\left\{\begin{array}{l}{∠B=∠DAF=45°}\\{BD=AD}\\{∠BDE=∠ADF}\end{array}\right.$，
$\therefore \triangle BED$≌$\triangle AFD\left(ASA\right)$，
$\therefore BE=AF=6cm$，
故答案为：$6cm$.

$\because \angle BAC=90^{\circ}$，$AB=AC$，$AD\bot BC$于点$D$，
$\therefore BD=AD$，$\angle B=\angle C=\angle BAD=\angle DAC=45^{\circ}$，
$\because AD\bot BC$于点$D$，$\angle EDF=90^{\circ}$，
$\therefore \angle BDE+\angle ADE=\angle ADE+\angle ADF=90^{\circ}$，
$\therefore \angle BDE=\angle ADF$，
在$\triangle BED$与$\triangle AFD$中，
$\left\{\begin{array}{l}{∠B=∠DAF=45°}\\{BD=AD}\\{∠BDE=∠ADF}\end{array}\right.$，
$\therefore \triangle BED$≌$\triangle AFD\left(ASA\right)$，
$\therefore BE=AF=6cm$，
故答案为：$6cm$.