jmc

Since $\sin (180^{\circ }-x)=\sin x$ and $\cos (180^{\circ }-x)=-\cos x$ for any angle, we have $$\tan (180^{\circ }-x)=\frac{\sin (180^{\circ }-x)}{\cos (180^{\circ }-x)}=\frac{\sin x}{-\cos x}=-\tan x$$for any angle for which $\tan x$ is defined. Therefore, $\tan \angle BDC=-\tan \angle BDA$.

From the Pythagorean Theorem, we have $AC=\sqrt{B{C}^2-B{A}^2}=24$. Since $D$ is the midpoint of $\overline{AC}$, we have $AD=AC/2=12$. Therefore, we have $\tan \angle BDC=-\tan \angle BDA=-\frac{BA}{AD}=\boxed{-\frac{7}{12}}$.