5. Given an acute triangle $\triangle A B C$ with unequal side lengths, $D$ is a point on side $B C$, and $\angle B A D+\angle C=90^{\circ}$. Then, $A D$ must pass through the ( ) of $\triangle A B C$.
(A) Circumcenter
(B) Incenter
(C) Centroid
(D) Orthocenter
5.A.

As shown in Figure 7, extend $AD$ to intersect the circumcircle $\odot O$ of $\triangle ABC$ at $E$, and connect $BE$. Then $\angle E = \angle C$.

Since $\angle BAD + \angle C = 90^{\circ}$, we have
$$
\begin{array}{l}
\angle BAD + \angle E = 90^{\circ}. \\
\text{Therefore, } \angle ABE = 90^{\circ}.
\end{array}
$$

Thus, $AD$ must pass through the circumcenter $O$ of $\triangle ABC$.