全等三角形的判定,如图,AG⊥BC,AB=AE,AC=AF,∠BAE=∠CAF=90°,过F、E分别作射线GA的垂线,垂足分别为PQ.①求证:EP=FQ②连接EF交射线GA于点H,求证:EH=FH
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![全等三角形的判定,如图,AG⊥BC,AB=AE,AC=AF,∠BAE=∠CAF=90°,过F、E分别作射线GA的垂线,垂足分别为PQ.①求证:EP=FQ②连接EF交射线GA于点H,求证:EH=FH](/uploads/image/z/8760234-66-4.jpg?t=%E5%85%A8%E7%AD%89%E4%B8%89%E8%A7%92%E5%BD%A2%E7%9A%84%E5%88%A4%E5%AE%9A%2C%E5%A6%82%E5%9B%BE%2CAG%E2%8A%A5BC%2CAB%3DAE%2CAC%3DAF%2C%E2%88%A0BAE%3D%E2%88%A0CAF%3D90%C2%B0%2C%E8%BF%87F%E3%80%81E%E5%88%86%E5%88%AB%E4%BD%9C%E5%B0%84%E7%BA%BFGA%E7%9A%84%E5%9E%82%E7%BA%BF%2C%E5%9E%82%E8%B6%B3%E5%88%86%E5%88%AB%E4%B8%BAPQ.%E2%91%A0%E6%B1%82%E8%AF%81%EF%BC%9AEP%3DFQ%E2%91%A1%E8%BF%9E%E6%8E%A5EF%E4%BA%A4%E5%B0%84%E7%BA%BFGA%E4%BA%8E%E7%82%B9H%2C%E6%B1%82%E8%AF%81%EF%BC%9AEH%3DFH)
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