Because $ABCD$ are concyclic, $\angle{A}+\angle{C}=180^\circ$, or $\cos\angle{A}=-\cos\angle{C}$. Applying Law of Cosines on $\triangle{ABD}$ and $\triangle{CBD}$, respectively: $$ \left\{ \begin{array}{c} BD^2 = AB^2 + AD^2 - 2AB\cdot AD\cos\angle{A}\\ BD^2 = CB^2 + CD^2 - 2CB\cdot CD\cos\angle{C} \end{array} \right. $$ Setting $\cos\angle{C}=\cos\angle{A}$ and all the given conditions leads to $$\cos\angle{A}=\frac{25^2+60^2-39^2-52^2}{2\times(25\times 60 + 29\times 52)}=0\implies \angle{A}=90^\circ$$ This means $BD$ is the diameter. It follows $BD=\sqrt{25^2+60^2}=65$. Hence the answer is $\boxed{65\pi}$.