Segmented

We clearly have the $7,24,25$ primitive Pythagorean triplet for $\triangle{ABC}.$ Let point $D$ divide hypothenuse $AC$ into lengths $AD= x, DC = 25-x$ such that:

$\frac{x}{25-x} = \frac{2}{3} \Rightarrow x = 10$

or $AD = 10, DC = 15.$ We can then compute $BD^{2}$ per the Law of Cosines. WLOG, let us utilize $\triangle{BCD}:$

$BD^2 = BC^2 + CD^2 - 2(BC)(CD)\cos \angle C$ ;

or $BD^2 = 24^2 + 15^2 - 2(24)(15) \cdot \cos(\arccos \frac{24}{25}) = 801 - \frac{3456}{5} = \boxed{\frac{549}{5}}.$