It is known that PA is perpendicular to the plane of circle O, AB is the diameter of circle O, C is any point on circle O, and AE is perpendicular to PC through a Results: AE was perpendicular to the plane of PBC

Connect AC and BC, BC ⊥ PA, BC ⊥ AC get BC ⊥ plane PAC, then AE ⊥ BC, and because AE ⊥ PC, so AE ⊥ plane PBC
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