如图一,已知在Rt△ABC中,∠ABC=90,∠CAB=30,BC=5.过点A作AE⊥AB,且AE=15,连接BE交AC于点P.(1)以点A为圆心,AP为半径作⊙A,试判断BE与⊙A是否相切,并说明理由;(2)如图二,在(1)的条件
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![如图一,已知在Rt△ABC中,∠ABC=90,∠CAB=30,BC=5.过点A作AE⊥AB,且AE=15,连接BE交AC于点P.(1)以点A为圆心,AP为半径作⊙A,试判断BE与⊙A是否相切,并说明理由;(2)如图二,在(1)的条件](/uploads/image/z/2559707-35-7.jpg?t=%E5%A6%82%E5%9B%BE%E4%B8%80%2C%E5%B7%B2%E7%9F%A5%E5%9C%A8Rt%E2%96%B3ABC%E4%B8%AD%2C%E2%88%A0ABC%3D90%26%2361616%3B%2C%E2%88%A0CAB%3D30%26%2361616%3B%2CBC%3D5.%E8%BF%87%E7%82%B9A%E4%BD%9CAE%E2%8A%A5AB%2C%E4%B8%94AE%3D15%2C%E8%BF%9E%E6%8E%A5BE%E4%BA%A4AC%E4%BA%8E%E7%82%B9P.%EF%BC%881%EF%BC%89%E4%BB%A5%E7%82%B9A%E4%B8%BA%E5%9C%86%E5%BF%83%2CAP%E4%B8%BA%E5%8D%8A%E5%BE%84%E4%BD%9C%E2%8A%99A%2C%E8%AF%95%E5%88%A4%E6%96%ADBE%E4%B8%8E%E2%8A%99A%E6%98%AF%E5%90%A6%E7%9B%B8%E5%88%87%2C%E5%B9%B6%E8%AF%B4%E6%98%8E%E7%90%86%E7%94%B1%EF%BC%9B%EF%BC%882%EF%BC%89%E5%A6%82%E5%9B%BE%E4%BA%8C%2C%E5%9C%A8%EF%BC%881%EF%BC%89%E7%9A%84%E6%9D%A1%E4%BB%B6)
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