如图,在等边三角形ABC的边AB上取一点P,使PB=2PA,过P分别作PD⊥BC于D,PE⊥AB且交AC于E,求证:PD=EP
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![如图,在等边三角形ABC的边AB上取一点P,使PB=2PA,过P分别作PD⊥BC于D,PE⊥AB且交AC于E,求证:PD=EP](/uploads/image/z/6961952-56-2.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E7%AD%89%E8%BE%B9%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E8%BE%B9AB%E4%B8%8A%E5%8F%96%E4%B8%80%E7%82%B9P%2C%E4%BD%BFPB%3D2PA%2C%E8%BF%87P%E5%88%86%E5%88%AB%E4%BD%9CPD%E2%8A%A5BC%E4%BA%8ED%2CPE%E2%8A%A5AB%E4%B8%94%E4%BA%A4AC%E4%BA%8EE%2C%E6%B1%82%E8%AF%81%EF%BC%9APD%3DEP)
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