Given an equilateral $\triangle ABC$, find the locus of points $P$ such that $\angle APB=\angle BPC$.
1. **Consider the case when \( P \) is inside the equilateral triangle \( \triangle ABC \):**

   - Since \( \triangle ABC \) is equilateral, each angle is \( 60^\circ \).
   - Given \( \angle APB = \angle BPC \), we can use the Law of Sines (LOS) in \( \triangle APB \) and \( \triangle BPC \):
     \[
     \sin(\angle BAP) = \sin(\angle BCP)
     \]
   - Since \( \angle BAP \) and \( \angle BCP \) are both acute, it follows that:
     \[
     \angle BAP = \angle BCP
     \]
   - Therefore, \( P \) lies on the angle bisector of \( \angle ABC \).

2. **Consider the case when \( P \) is outside the triangle:**

   - If \( \angle PAC \leq 120^\circ \) and \( \angle PCA \leq 120^\circ \), then:
     \[
     \sin(\angle BAP) = \sin(\angle BCP)
     \]
     - If \( \angle BAP = \angle BCP \), then \( P \) lies on the angle bisector of \( \angle ABC \).
     - If \( \angle BAP + \angle BCP = 180^\circ \), then \( ABCP \) is cyclic, and \( \angle APC \) lies on the minor arc \( \overarc{AC} \) of the circumcircle of \( \triangle ABC \).

   - If \( \angle PCA \geq 120^\circ \) or \( \angle PCB \geq 120^\circ \) or \( \angle PBC \leq 120^\circ \) or \( \angle PAB \leq 120^\circ \) or \( \angle PAC \geq 120^\circ \), then:
     \[
     \angle APC = 0^\circ
     \]
     - Thus, \( P \) lies on the line segment \( \overline{AC} \).

   - If \( \angle PBA \geq 120^\circ \) and \( \angle PBC \geq 120^\circ \), then \( P \) lies on the angle bisector of \( \angle ABC \).

3. **Verification of all points on the locus:**

   - All points \( P \) lying on the angle bisector of \( \angle ABC \) satisfy \( \angle APB = \angle BPC \).
   - For all points on the minor arc \( \overarc{AC} \) of the circumcircle of \( \triangle ABC \), we have \( \angle APB = \angle BPC = 60^\circ \) by the inscribed angle theorem.
   - All points \( P \) lying on \( \overline{AC} \) but not on the segment \( AC \) also satisfy the condition.

The final answer is the locus of points \( \boxed{ P } \) such that \( \angle APB = \angle BPC \) is the union of the angle bisector of \( \angle ABC \), the minor arc \( \overarc{AC} \) of the circumcircle of \( \triangle ABC \), and the line segment \( \overline{AC} \) but not including the segment \( AC \).