14. Let $M$ and $N$ be the midpoints of $A B$ and $C D$, and let $M^{\prime}, N^{\prime}$ be their projections on $C D$ and $A B$, respectively. We know that $M M^{\prime}=A B / 2$, and hence
$$S_{A B C D}=S_{A M D}+S_{B M C}+S_{C M D}=\frac{1}{2}\left(S_{A B D}+S_{A B C}\right)+\frac{1}{4} A B \cdot C D$$
The line $A B$ is tangent to the circle with diameter $C D$ if and only if $N N^{\prime}=C D / 2$, or equivalently,
$$S_{A B C D}=S_{A N D}+S_{B N C}+S_{A N B}=\frac{1}{2}\left(S_{B C D}+S_{A C D}\right)+\frac{1}{4} A B \cdot C D$$
By (1), this is further equivalent to $S_{A B C}+S_{A B D}=S_{B C D}+S_{A C D}$. But since $S_{A B C}+S_{A C D}=S_{A B D}+S_{B C D}=S_{A B C D}$, this reduces to $S_{A B C}=S_{B C D}$, i.e., to $B C \| A D$.