Example 4 $O$ is a fixed point on the plane, $A, B, C$ are three non-collinear points on the plane, and the moving point $P$ satisfies $\overrightarrow{O P}=\overrightarrow{O A}+$ $\lambda\left(\frac{\overrightarrow{A B}}{|\overrightarrow{A B}|}+\frac{\overrightarrow{A C}}{|\overrightarrow{A C}|}\right), \lambda \in[0,+\infty)$, then the trajectory of $P$ must pass through the
A. Circumcenter
B. Incenter
C. Centroid
D. Orthocenter
Analyze $\frac{\overrightarrow{A B}}{|\overrightarrow{A B}|}$ is a unit vector in the same direction as $\overrightarrow{A B}$. $\frac{\overrightarrow{A C}}{|\overrightarrow{A C}|}$ is a unit vector in the same direction as $\overrightarrow{A C}$, with $A$ as the common starting point, and using these two unit vectors as adjacent sides to form a rhombus $A B_{0} P_{0} C_{0}$ (as shown in the figure), then their sum vector $\frac{\overrightarrow{A B}}{|\overrightarrow{A B}|}+\frac{\overrightarrow{A C}}{|\overrightarrow{A C}|}$ is the vector $\overrightarrow{A P_{0}}$ determined by the diagonal of the rhombus with $A$ as the starting point. At the same time, knowing that the diagonals of a rhombus bisect a pair of opposite angles, $A P_{0}$ bisects $\angle B A C$. Also, $\lambda\left(\frac{\overrightarrow{A B}}{|\overrightarrow{A B}|}+\frac{\overrightarrow{A C}}{|\overrightarrow{A C}|}\right)=\lambda \overrightarrow{A P_{0}}(\lambda \geqslant 0)$ is a vector with $A$ as the starting point and in the same direction as $A P_{0}$. $\overrightarrow{A P}=\overrightarrow{O P}-\overrightarrow{O A}=\lambda \overrightarrow{A P_{0}}(\lambda \geqslant 0)$, hence the trajectory of $P$ is the angle bisector of $\angle B A C$ (including point $A$), so the answer is B.