Suppose first that $D$ lies between $B$ and $C$. Let $A B C$ be inscribed in circle $\omega$, and extend $A D$ to intersect $\omega$ again at $E$. Note that $A$ subtends a quarter of the circle, so in particular, the chord through $C$ perpendicular to $B C$ and parallel to $A D$ has length $B C=5$. Therefore, $A D=5+D E$. By power of a point, $6=B D \cdot D C=A D \cdot D E=$ $A D^{2}-5 A D$, implying $A D=6$, so the area of $A B C$ is $\frac{1}{2} B C \cdot A D=15$. If $D$ does not lie between $B$ and $C$, then $B C=1$, so $A$ lies on a circle of radius $\sqrt{2} / 2$ through $B$ and $C$. But then it is easy to check that the perpendicular to $B C$ through $D$ cannot intersect the circle, a contradiction.