Let $ABCD$ be a cyclic quadrilateral. The side lengths of $ABCD$ are distinct integers less than $15$ such that $BC\cdot CD=AB\cdot DA$. What is the largest possible value of $BD$?
$\textbf{(A)}\ \sqrt{\dfrac{325}{2}} \qquad \textbf{(B)}\ \sqrt{185} \qquad \textbf{(C)}\ \sqrt{\dfrac{389}{2}} \qquad \textbf{(D)}\ \sqrt{\dfrac{425}{2}} \qquad \textbf{(E)}\ \sqrt{\dfrac{533}{2}}$
Let $AB = a$, $BC = b$, $CD = c$, and $AD = d$.  We see that by the Law of Cosines on $\triangle ABD$ and $\triangle CBD$, we have:
$BD^2 = a^2 + d^2 - 2ad\cos{\angle BAD}$.  
$BD^2 = b^2 + c^2 - 2bc\cos{\angle BCD}$.
We are given that $ad = bc$ and $ABCD$ is a cyclic quadrilateral. As a property of cyclic quadrilaterals, opposite angles are supplementary so $\angle BAD = 180 - \angle BCD$, therefore $\cos{\angle BAD} = -\cos{\angle BCD}$. So, $2ad\cos{\angle BAD} = -2bc\cos{\angle BCD}$. 
Adding, we get $2BD^2 = a^2 + b^2 + c^2 + d^2$.
We now look at the equation $ad = bc$.  Suppose that $a = 14$.  Then, we must have either $b$ or $c$ equal $7$.  Suppose that $b = 7$.  We let $d = 6$ and $c = 12$.  
$2BD^2 = 196 + 49 + 36 + 144 = 425$, so our answer is $\boxed{\textbf{(D)} \sqrt{\frac{425}{2}}}$.