In planar quadrilateral $A B C D$, $\angle A D C = 90 ^ { \circ }$, $\angle A = 45 ^ { \circ }$, $A B = 2$, $B D = 5$. (1) Find $\cos \angle A D B$; (2) If $D C = 2 \sqrt { 2 }$, find $B C$.
(1) In $\triangle A B D$, by the sine rule we have $\frac { B D } { \sin \angle A } = \frac { A B } { \sin \angle A D B }$. From the given conditions, $\frac { 5 } { \sin 45 ^ { \circ } } = \frac { 2 } { \sin \angle A D B }$, so $\sin \angle A D B = \frac { \sqrt { 2 } } { 5 }$. From the given conditions, $\angle A D B < 90 ^ { \circ }$, so $\cos \angle A D B = \sqrt { 1 - \frac { 2 } { 25 } } = \frac { \sqrt { 23 } } { 5 }$. (2) From the given conditions and (1), $\cos \angle B D C = \sin \angle A D B = \frac { \sqrt { 2 } } { 5 }$. In $\triangle B C D$, by the cosine rule: $B C ^ { 2 } = B D ^ { 2 } + D C ^ { 2 } - 2 \cdot B D \cdot D C \cdot \cos \angle B D C = 25 + 8 - 2 \times 5 \times 2 \sqrt { 2 } \times \frac { \sqrt { 2 } } { 5 } = 25$. So $B C = 5$.
In planar quadrilateral $A B C D$, $\angle A D C = 90 ^ { \circ }$, $\angle A = 45 ^ { \circ }$, $A B = 2$, $B D = 5$.
(1) Find $\cos \angle A D B$;
(2) If $D C = 2 \sqrt { 2 }$, find $B C$.