Let $ABC$ be a triangle and $I$ the center of its incircle. $P$ is a point inside $ABC$ such that $\angle PBA +\angle PCA = \angle PBC + \angle PCB$. Prove that $AP\geq AI$ with equality iff $P=I$.
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Tags: geometry, geometry unsolved
Let $ABC$ be a triangle and $I$ the center of its incircle. $P$ is a point inside $ABC$ such that $\angle PBA +\angle PCA = \angle PBC + \angle PCB$. Prove that $AP\geq AI$ with equality iff $P=I$.