Proof: extending AG to BC is q point, connecting MQ. AQ is perpendicular to BC because AB=AC and MB=MC because the projection of point P on the plane ABC is just the center of gravity G, which is perpendicular to BC. Because AQ belongs to PAG plane, MQ belongs to PAG plane, and BC belongs to both PAG plane and BCM plane, PAG plane is perpendicular to BCM plane.

Solution: from the original problem, we can know that P is the projection of the center of gravity G, so the distance from P to A, B and C is equal, which is MQ vertical. When m is the midpoint of pa, connect PQ, and the triangle PQM = root 1 1/2 root 10= root1/0/20.