$\textrm{Vì ΔABC vuông cân tại A}$
$→ \widehat{B}=\widehat{C} = 45^{0}$
$\textrm{Ta có: DBA+ABC=$180^{0}$ (kề bù)}$
$→ \widehat{ABD}=180^{0}-45^{0}=135^{0}$
$\textrm{Ta có: ΔBAD cân tại B}$
$→ \widehat{D}+\widehat{B}+\widehat{A}=180^{0}$
$→ 2D=180^{0}-135^{0}$
$→ 2D=45^{0}$
$→ D=\frac{45}{2}$
$→ D=22,5^{0}$
$\textrm{Vậy $\widehat{ADB}=22,5^{0}$}$