25. In isosceles $\triangle A B C$, $A B=A C, A D \perp B C$ at point $D, B C=16, A D=15$, and a semicircle $\odot D$ is tangent to the two legs. Then the radius of the semicircle $\odot D$ is ( ).
(A) $\frac{17 \sqrt{2}}{2}$
(B) $\frac{120}{17}$
(C) $\frac{17 \sqrt{3}}{2}$
(D) $4 \sqrt{3}$
(E) 10
25. B.

As shown in Figure 4, let the semicircle $\odot D$ be tangent to $AB$ at point $E$. Then $DE \perp AB$.
From the given information, we have $BD = DC$.
Also, $BC = 16$, so
$$
\begin{array}{l}
BD = 8 \Rightarrow AB = \sqrt{BD^2 + AD^2} = 17 . \\
\text{By } S_{\triangle ABD} = \frac{1}{2} AD \cdot BD = \frac{1}{2} AB \cdot DE \\
\Rightarrow DE = \frac{AD \cdot BD}{AB} = \frac{15 \times 8}{17} = \frac{120}{17} .
\end{array}
$$