如图,在△ABC中,∠ABC和∠ACB的平分线BE、CF相交于点P.
(1)若∠ABC=70,∠ACB=50,则∠BPC= ;
(2)求证:∠BPC=180﹣(∠ABC+∠ACB);
解(1)PBC+∠PCB= (∠ABC+∠ACB)= ×120°=60°,
在△PBC中,∠BPC=180°−(∠PBC+∠PCB)=180°−60°=120°
(2)证明:∵∠ABC和∠ACB的平分线BE、CF相交于点P,
∴∠PBC=∠ABC, ∠PCB=∠ACB,
∵∠BPC +∠PBC+∠PCB=180°,
∴∠BPC=180°-(∠PBC+∠PCB)= 180°-(∠ABC +∠ACB) =180°- (∠ABC+∠ACB) ∴∠BPC=180°- (∠ABC+∠ACB);