设$\angle B=\angle C=y$，$\angle EDC=x$，
$\because \angle BAC=56^{\circ}$，$AB=AC$，
$\therefore \angle B=\angle C=\frac{1}{2}(180^{\circ}-\angle BAC)=62^{\circ}$，
$\angle EDC+\angle C=x+y=\angle AED$，
$\because AD=AE$，
$\therefore \angle ADE=\angle AED=x+y$，
$\therefore \angle ADC=\angle ADE+\angle EDC=2x+y$，
$\because \angle ADC=\angle B+\angle BAD$，
$\therefore 2x+y=y+32$，
$\therefore x=16$，
$\therefore \angle EDC$的度数是$16^{\circ}$.
故答案为：$16$.

设$\angle B=\angle C=y$，$\angle EDC=x$，
$\because \angle BAC=56^{\circ}$，$AB=AC$，
$\therefore \angle B=\angle C=\frac{1}{2}(180^{\circ}-\angle BAC)=62^{\circ}$，
$\angle EDC+\angle C=x+y=\angle AED$，
$\because AD=AE$，
$\therefore \angle ADE=\angle AED=x+y$，
$\therefore \angle ADC=\angle ADE+\angle EDC=2x+y$，
$\because \angle ADC=\angle B+\angle BAD$，
$\therefore 2x+y=y+32$，
$\therefore x=16$，
$\therefore \angle EDC$的度数是$16^{\circ}$.
故答案为：$16$.