如图所示,等边三角形ABC的边长是6,点P在边AB上,点Q在BC的延长线上,且AP=CQ,设PQ与AC相交于点D.(1)当∠DQC=30°时,求AP的长(2)作PE⊥AC与E,求证:DE=AE+CD
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![如图所示,等边三角形ABC的边长是6,点P在边AB上,点Q在BC的延长线上,且AP=CQ,设PQ与AC相交于点D.(1)当∠DQC=30°时,求AP的长(2)作PE⊥AC与E,求证:DE=AE+CD](/uploads/image/z/9865606-22-6.jpg?t=%E5%A6%82%E5%9B%BE%E6%89%80%E7%A4%BA%2C%E7%AD%89%E8%BE%B9%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E8%BE%B9%E9%95%BF%E6%98%AF6%2C%E7%82%B9P%E5%9C%A8%E8%BE%B9AB%E4%B8%8A%2C%E7%82%B9Q%E5%9C%A8BC%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%B8%8A%2C%E4%B8%94AP%3DCQ%2C%E8%AE%BEPQ%E4%B8%8EAC%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9D.%EF%BC%881%EF%BC%89%E5%BD%93%E2%88%A0DQC%3D30%C2%B0%E6%97%B6%2C%E6%B1%82AP%E7%9A%84%E9%95%BF%EF%BC%882%EF%BC%89%E4%BD%9CPE%E2%8A%A5AC%E4%B8%8EE%2C%E6%B1%82%E8%AF%81%EF%BC%9ADE%3DAE%2BCD)
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