连接$ME$、$NE$，如图所示：

$\because AD\bot BC$，$BE\bot AC$，
$\therefore \angle ADB=\angle ADC=\angle BEC=\angle BEA=90^{\circ}$，
$\because \angle EBC+\angle BCE=\angle EAH+\angle BCE=90^{\circ}$，
$\therefore \angle EBC=\angle EAH$，
$\because \angle BAC=45^{\circ}$，
$\therefore \angle ABE=45^{\circ}$，
$\therefore AE=BE$，
$\therefore \triangle AHE$≌$\triangle BCE\left(ASA\right)$，
$\therefore AH=BC$，
$\because M$，$N$分别为$AH$，$BC$的中点，
$\therefore ME=\frac{1}{2}AH=AM，NE=\frac{1}{2}BC=CN$，
$\therefore ME=NE$，$\angle MEA=\angle MAE$，$\angle NEC=\angle NCE$，
$\therefore \angle MEA+\angle NEC=\angle MAE+\angle NCE=90^{\circ}$，
$\therefore \angle MEN=90^{\circ}$，
$\therefore \triangle MEN$为等腰直角三角形，
$\because MN=\sqrt{2}$，$ME=NE$，
$\therefore NE=\sqrt{M{N}^{2}-M{E}^{2}}=1$，
$\therefore BC=2NE=2$.
故答案为：$2$.

连接$ME$、$NE$，如图所示：

$\because AD\bot BC$，$BE\bot AC$，
$\therefore \angle ADB=\angle ADC=\angle BEC=\angle BEA=90^{\circ}$，
$\because \angle EBC+\angle BCE=\angle EAH+\angle BCE=90^{\circ}$，
$\therefore \angle EBC=\angle EAH$，
$\because \angle BAC=45^{\circ}$，
$\therefore \angle ABE=45^{\circ}$，
$\therefore AE=BE$，
$\therefore \triangle AHE$≌$\triangle BCE\left(ASA\right)$，
$\therefore AH=BC$，
$\because M$，$N$分别为$AH$，$BC$的中点，
$\therefore ME=\frac{1}{2}AH=AM，NE=\frac{1}{2}BC=CN$，
$\therefore ME=NE$，$\angle MEA=\angle MAE$，$\angle NEC=\angle NCE$，
$\therefore \angle MEA+\angle NEC=\angle MAE+\angle NCE=90^{\circ}$，
$\therefore \angle MEN=90^{\circ}$，
$\therefore \triangle MEN$为等腰直角三角形，
$\because MN=\sqrt{2}$，$ME=NE$，
$\therefore NE=\sqrt{M{N}^{2}-M{E}^{2}}=1$，
$\therefore BC=2NE=2$.
故答案为：$2$.