jmc

Since $\cos (180^{\circ }-x)=-\cos x$ for any angle, we have $\cos \angle BDC=-\cos \angle BDA$.

From the Pythagorean Theorem, we have $AC=\sqrt{B{C}^2-B{A}^2}=9$. Applying the Angle Bisector Theorem to $\overline{BD}$, we have $\frac{AD}{DC}=\frac{AB}{BC}=\frac{4}{5}$. Since $AD+DC=AC=9$ and $\frac{AD}{DC}=\frac{4}{5}$, we have $AD=4$ and $DC=5$.

Applying the Pythagorean Theorem to $△ABD$ gives $BD=\sqrt{A{B}^2+A{D}^2}=\sqrt{144+16}=4\sqrt{10}$, so $$\cos BDC=-\cos BDA=-\frac{AD}{BD}=-\frac{4}{4\sqrt{10}}=-\frac{1}{\sqrt{10}}=\boxed{-\frac{\sqrt{10}}{10}}.$$