Let p be a point outside the plane where △ ABC lies, the distance between P and a, B, C is equal, and ∠ BAC is a right angle

Prove: as shown in the answer figure, take the midpoint D of BC, connect PD and ad, ∵ D is the midpoint of the hypotenuse BC of the right triangle ABC, ∵ BD = CD = ad, PA = Pb = PC, PD is the common edge ∵ PDA = ∵ PDB = ∵ PDC = 90 °, PD ⊥ BC, PD ⊥ Da, ∵ PD ⊥ plane ABC ∵ and PD ⊂ plane PCB ⊥ plane ABC

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