Plane PAB perpendicular to plane ABC
Pg is on the plane.
In addition, the plane PAC is perpendicular to the plane ABC.
Pg is on the plane.
There can only be one intersection between two planes.
So point g and point a coincide.
That is, PA is perpendicular to the plane ABC.
(2) The first is the position of point E..
E is the vertical foot of the triangle PBC.
E is on one side of the triangle.
Suppose e connects AE and PE on BC side.
Then there is PE⊥BC AE⊥ People's Bank.
∴PE⊥AE ∵PE⊥BC
The coincidence of ABC E and point A on the ∴PE⊥ plane does not meet the requirements.
Suppose e is connected to AE, on the edge PC.
Then BE⊥PC AE⊥ people's bank.
∴AE⊥BE AE⊥BC
∴BE⊥ noodle bag
You ∵PA⊥ noodles ABC
∴PA⊥BC and ∵AE⊥BC
∴BC⊥ ∴ Noodles Bao
That is to say, point C coincides with point E ∠ ACB = 90.
Similarly, suppose that when E is on PB∠ABC = 90°, point E coincides with point B.
To sum up, triangle ABC is a right triangle.