19.
(1)
By the sine rule:
$$B D = \frac { a \sin C } { \sin B } = \frac { a c } { b } = \frac { b ^ { 2 } } { b } = b$$
(2)
Note that $\angle B D A$ and $\angle B D C$ are supplementary, so their cosines are opposite in sign. Applying the cosine rule in triangles $\triangle B D A$ and $\triangle B D C$:
$$\cos \angle B D A + \cos \angle B D C = \frac { b ^ { 2 } + \left( \frac { 2 } { 3 } b \right) ^ { 2 } - c ^ { 2 } } { 2 \cdot \left( \frac { 2 } { 3 } b \right) \cdot b } + \frac { b ^ { 2 } + \left( \frac { 1 } { 3 } b \right) ^ { 2 } - a ^