Inradius

From $25^2 + 60^2 = 65^2$ and $39^2+52^2=65^2$ and by Pythagorean theorem , we have $\angle B = \angle D = 90^\circ$ . Since the sum of opposite angles of the quadrilateral $ABCD$ is $180^\circ$ , $ABCD$ is a cyclic quadrilateral . By Ptolemy's theorem :

$\begin{aligned} AC \cdot BD & = AB \cdot CD + BC \cdot DA \\ 65 \cdot BD & = 25 \cdot 39 + 60 \cdot 52 \\ \implies BD & = \frac {25 \cdot 39 + 60 \cdot 52}{65} = 63 \end{aligned}$

Let the inradius of $\triangle BCD$ be $r$ , then $\Delta = s r$ , where $\Delta$ and $s=\dfrac {60+39+63}2=81$ are the area and semiperimeter of $\triangle BCD$ . By Heron's formula , we have $\Delta = \sqrt{s(s-60)(s-39)(s-63)}=1134$ . Therefore $r = \dfrac {1134}{81} = \boxed{14}$ .