8. Let $D$ be a point inside $\triangle A B C$ such that $\angle B A D=$ $\angle B C D$, and $\angle B D C=90^{\circ}$. Given $A B=5, B C=$ $6, M$ is the midpoint of $A C$. Then $D M=$ $\qquad$
8. $\frac{\sqrt{11}}{2}$.

As shown in Figure 5, extend $CD$ to point $E$ such that $DE = DC$, and connect $BE$.
Then, $\triangle BDE \cong \triangle BDC$.
Therefore, $BE = BC = 6$, $\angle BED = \angle BCD = \angle BAD$.
Thus, points $A$, $D$, $B$, and $E$ are concyclic.
Hence, $\angle EAB = \angle EDB = 90^{\circ}$.
So, $AE = \sqrt{6^2 - 5^2} = \sqrt{11}$.
Therefore, $DM = \frac{\sqrt{11}}{2}$.