5. As shown in Figure $1, M$ is the midpoint of side $CD$ of the inscribed quadrilateral $ABCD$ in the semicircle $\odot O$ with diameter $AB$, $MN \perp AB$ at point $N$, $AB=10, AD=AN=$ 3. Then $BC=(\quad)$.
(A) 5
(B) 6
(C) 7
(D) 8
5. C.

As shown in Figure 5, draw perpendiculars from points $C$ and $D$ to $AB$, with the feet of the perpendiculars being $E$ and $F$, respectively. Connect $AC$ and $BD$.
Let $AF = x$. Then
$$
\begin{array}{c}
E N = F N = 3 - x, \\
B E = B N - E N = (10 - 3) - (3 - x) = 4 + x .
\end{array}
$$

In right triangles $\triangle ABD$ and $\triangle ABC$, by the projection theorem, we have
$$
\begin{array}{l}
A D^{2} = A F \cdot A B, \quad B C^{2} = B E \cdot A B . \\
\text { Therefore, } x = A F = \frac{A D^{2}}{A B} = \frac{9}{10}, \\
B C^{2} = \left(4 + \frac{9}{10}\right) \times 10 = 49, \quad B C = 7 .
\end{array}
$$