平面几何竞赛题设P为△ABC内的一点,且∠APB=∠BPC=∠CPA=120°,O为异于P的任意一点,求证:OA+OB+OC>PA+PB+PC
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![平面几何竞赛题设P为△ABC内的一点,且∠APB=∠BPC=∠CPA=120°,O为异于P的任意一点,求证:OA+OB+OC>PA+PB+PC](/uploads/image/z/10725912-0-2.jpg?t=%E5%B9%B3%E9%9D%A2%E5%87%A0%E4%BD%95%E7%AB%9E%E8%B5%9B%E9%A2%98%E8%AE%BEP%E4%B8%BA%E2%96%B3ABC%E5%86%85%E7%9A%84%E4%B8%80%E7%82%B9%2C%E4%B8%94%E2%88%A0APB%3D%E2%88%A0BPC%3D%E2%88%A0CPA%3D120%C2%B0%2CO%E4%B8%BA%E5%BC%82%E4%BA%8EP%E7%9A%84%E4%BB%BB%E6%84%8F%E4%B8%80%E7%82%B9%2C%E6%B1%82%E8%AF%81%EF%BC%9AOA%2BOB%2BOC%EF%BC%9EPA%2BPB%2BPC)
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