In tetrahedral PABC, PA, Pb and PC are perpendicular. It is proved that △ ABC is an acute triangle In tetrahedral PABC, PA, Pb and PC are perpendicular. It is proved that ① △ ABC is an acute triangle. ② the square of s △ ABC = the square of s △ PBC + the square of s △ PAB + the square of s △ PCA

In tetrahedral PABC, PA, Pb and PC are perpendicular. It is proved that △ ABC is an acute triangle In tetrahedral PABC, PA, Pb and PC are perpendicular. It is proved that ① △ ABC is an acute triangle. ② the square of s △ ABC = the square of s △ PBC + the square of s △ PAB + the square of s △ PCA

① Let H be the perpendicularity of △ ABC. Prove: ∩ PA ^ pbpa ^ PC and Pb ∩ PC = P ∩ PA ^ side PBC ∩ bc plane PBD ∩ PA ^ BC ∩ h is the perpendicularity of △ ABC ∩ ah ^ BC ∩ PA ∩ ah = a ∩ BC ^ section PAH and pH plane PAH ∩ BC ^ pH. similarly, prove: ab ^ pH and ABC = B ∩ pH ^ plane ABC, let the intersection of ah and straight line BC