如图所示,点P是△ABC内一点,PD⊥BC于点D,PE⊥AC于点E,PF⊥AB于点F,PD=PE=PF.若∠A=70°,∠BPC等于

Question

如图所示,点P是△ABC内一点,PD⊥BC于点D,PE⊥AC于点E,PF⊥AB于点F,PD=PE=PF.若∠A=70°,∠BPC等于

Answer 1

∠A=70° ∴∠ABC＋∠ACB=110°
∵PD⊥BC于点D,PE⊥AC于点E,PF⊥AB于点F,PD=PE=PF
∴BP平分∠ABC,CP平分∠ACB
∴∠PBC=½∠ABC,∠PCB=½∠ACB
∴∠PBC＋∠PCB=½∠ABC＋½∠ACB=½﹙∠ABC＋∠ACB﹚=55°
∴∠BPC=125°