如图,△ABC是等边三角形,且BM=CN,AM与BN交于点P,NH⊥AM于H,求证:PN=2PH.
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![如图,△ABC是等边三角形,且BM=CN,AM与BN交于点P,NH⊥AM于H,求证:PN=2PH.](/uploads/image/z/10716883-43-3.jpg?t=%E5%A6%82%E5%9B%BE%2C%E2%96%B3ABC%E6%98%AF%E7%AD%89%E8%BE%B9%E4%B8%89%E8%A7%92%E5%BD%A2%2C%E4%B8%94BM%3DCN%2CAM%E4%B8%8EBN%E4%BA%A4%E4%BA%8E%E7%82%B9P%2CNH%E2%8A%A5AM%E4%BA%8EH%2C%E6%B1%82%E8%AF%81%EF%BC%9APN%3D2PH.)
如图,△ABC是等边三角形,且BM=CN,AM与BN交于点P,NH⊥AM于H,求证:PN=2PH.
如图,△ABC是等边三角形,且BM=CN,AM与BN交于点P,NH⊥AM于H,求证:PN=2PH.
如图,△ABC是等边三角形,且BM=CN,AM与BN交于点P,NH⊥AM于H,求证:PN=2PH.
CN=BM,CB=BA,