解:
(1)∵∠B和∠C的外角平分线相交于点P,
∴∠PBC=1/2∠EBC, ∠PCB=1/2∠FCB.
∵∠A=80°,∴∠ABC+∠ACB= 100°.
又∵∠ABC +∠ACB+∠EBC+∠FCB=360°,
∴∠EBC+∠FCB= 360°.
又∵∠PBC+∠PCB+∠BPC= 180°.
∴∠BPC = 180° - (∠PBC +∠PCB ) = 180°-1/2 (∠EBC+∠FCB)=50°
追问为什么我算出来等于55度?第二题呢?
追答∠P=90-1/2∠A
∠ACB=2b-a