三棱柱ABC-A1B1C1中,AB⊥AC,顶点A1在地面ABC上的射影恰为点B,且AB=AC=A1B=2 证明:平面A1AC⊥平面AB1B
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![三棱柱ABC-A1B1C1中,AB⊥AC,顶点A1在地面ABC上的射影恰为点B,且AB=AC=A1B=2 证明:平面A1AC⊥平面AB1B](/uploads/image/z/3745745-17-5.jpg?t=%E4%B8%89%E6%A3%B1%E6%9F%B1ABC-A1B1C1%E4%B8%AD%2CAB%E2%8A%A5AC%2C%E9%A1%B6%E7%82%B9A1%E5%9C%A8%E5%9C%B0%E9%9D%A2ABC%E4%B8%8A%E7%9A%84%E5%B0%84%E5%BD%B1%E6%81%B0%E4%B8%BA%E7%82%B9B%2C%E4%B8%94AB%3DAC%3DA1B%3D2+%E8%AF%81%E6%98%8E%3A%E5%B9%B3%E9%9D%A2A1AC%E2%8A%A5%E5%B9%B3%E9%9D%A2AB1B)
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