Problem: Let $ABC$ be a triangle with incenter $I$.
A point $P$ in the interior of the triangle satisfies
\[ \angle PBA + \angle PCA = \angle PBC + \angle PCB. \]
Show that $AP \ge AI$ and that equality holds if and only if $P=I$.

Solution: The condition rewrites as
\[
  \angle PBC + \angle PCB
  = (\angle B - \angle PBC)
  + (\angle C - \angle PCB)
  \implies
  \angle PBC + \angle PCB = \frac{\angle B + \angle C}{2}
\]
which means that
\[ \angle BPC = 180\dg - \frac{\angle B + \angle C}{2}
  = 90\dg + \frac{\angle A}{2}
  = \angle BIC.
\]
Since $P$ and $I$ are both inside $\triangle ABC$
that implies $P$ lies on the circumcircle of $\triangle BIC$.

It's well-known (by ``Fact 5'') that the circumcenter
of $\triangle BIC$ is the arc midpoint $M$ of $\widehat{BC}$.
Therefore
\[ AI + IM = AM \le AP + PM \implies AI \le AP \]
with equality holding iff $A$, $P$, $M$ are collinear, or $P=I$.