$\because$分别过点$A$、$B$作$AB$的垂线$AD$、$BC$，
$\therefore \angle A=\angle B=90^{\circ}$，
$\therefore \angle ADE+\angle AED=90^{\circ}$
$\because DE\bot CE$，
$\therefore \angle DEC=90^{\circ}$，
$\therefore \angle AED+\angle BEC=90^{\circ}$
$\therefore \angle ADE=\angle BEC$，
在$\triangle ADE$和$\triangle BEC$中，
$\left\{\begin{array}{l}{∠A=∠B}\\{∠ADE=∠BEC}\\{DE=EC}\end{array}\right.$，
$\therefore \triangle ADE$≌$\triangle BEC\left(AAS\right)$，
$\therefore AD=BE$，$AE=BC$，
$\because AD=AF$，
$\therefore BE=AF$，
$\therefore BF+EF=AE+EF$，
$\therefore BF=AE$，
$\therefore BF=BC$，
$\therefore \triangle BCF$是等腰直角三角形，
$\therefore \angle BFC=45^{\circ}$，
故答案为：$45^{\circ}$.

$\because$分别过点$A$、$B$作$AB$的垂线$AD$、$BC$，
$\therefore \angle A=\angle B=90^{\circ}$，
$\therefore \angle ADE+\angle AED=90^{\circ}$
$\because DE\bot CE$，
$\therefore \angle DEC=90^{\circ}$，
$\therefore \angle AED+\angle BEC=90^{\circ}$
$\therefore \angle ADE=\angle BEC$，
在$\triangle ADE$和$\triangle BEC$中，
$\left\{\begin{array}{l}{∠A=∠B}\\{∠ADE=∠BEC}\\{DE=EC}\end{array}\right.$，
$\therefore \triangle ADE$≌$\triangle BEC\left(AAS\right)$，
$\therefore AD=BE$，$AE=BC$，
$\because AD=AF$，
$\therefore BE=AF$，
$\therefore BF+EF=AE+EF$，
$\therefore BF=AE$，
$\therefore BF=BC$，
$\therefore \triangle BCF$是等腰直角三角形，
$\therefore \angle BFC=45^{\circ}$，
故答案为：$45^{\circ}$.