If P is a point outside the plane where △ ABC is located and the lines AP, BP and CP are perpendicular, then the projection of P on plane ABC is ()

orthocenter.
Let pH ⊥ plane ABC and H be perpendicular feet. Since the straight lines AP, BP and CP are perpendicular, so AP ⊥ plane PBC
So AP ⊥ BC, in the bottom surface ABC, it is easy to know that ah is the projection of PA. from the three perpendicular theorem, ah ⊥ BC,
Similarly, BH ⊥ AC, CH ⊥ AB, that is, h is perpendicular

If P is a point outside the plane where △ ABC is located and the lines AP, BP and CP are perpendicular, then the projection of P on plane ABC is ()

If P is a point outside the plane where △ ABC is located and the lines AP, BP and CP are perpendicular, then the projection of P on plane ABC is ()

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