14. As shown in the figure, $D$ is a point inside $\triangle A B C$, and it satisfies $A B=C D=4$, $\angle A+\angle B D C=180^{\circ}$. Determine the maximum value of $S_{\triangle A B C}-S_{\triangle B D C}$.
As shown in the figure, construct the symmetric point $D^{\prime}$ of $D$ with respect to $BC$. From $\angle A + \angle B D^{\prime} C = 180^{\circ} \Rightarrow A, B, C, D^{\prime}$ are concyclic, and $AB = CD^{\prime} = 4 \Rightarrow AD^{\prime} \parallel BC \Rightarrow ABCD^{\prime}$ is an isosceles trapezoid.
Construct $BE \parallel CD^{\prime}$ intersecting $AC$ at point $E$, then $BD^{\prime}CE$ is a parallelogram $\Rightarrow S_{\triangle BDC} = S_{\triangle BD^{\prime}C} = S_{\triangle BCE}$,
Thus, $S_{\triangle ABC} - S_{\triangle BDC} = S_{\triangle ABC} - S_{\triangle BCE} = S_{\triangle ABE}$. And $S_{\triangle ABE} = \frac{1}{2} AB^2 \sin \angle ABE \leqslant 8$, equality holds when $\angle ABE = 90^{\circ} \Rightarrow \angle BAC = 45^{\circ}$. Therefore, the maximum value of $S_{\triangle ABC} - S_{\triangle BDC}$ is 8.