smc

Let $AB=a$, $BC=b$, $CD=c$, and $AD=d$. We see that by the Law of Cosines on $△ABD$ and $△CBD$, we have: $B{D}^2=a^2+d^2-2ad\cos \angle BAD$. $B{D}^2=b^2+c^2-2bc\cos \angle BCD$. 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 $2B{D}^2=a^2+b^2+c^2+d^2$. 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 $c=12$. $2B{D}^2=196+49+36+144=425$, so our answer is $\boxed{\sqrt{\frac{425}{2}}}$.