Question

如图,已知△ABC中,∠B=80°(1)若点P是△ABC的内心,求∠P的度数 (2)点Q是∠A,∠C的外角平分线的交点,

Answer 1

(1)可得∠BPC=90°-∠A=50°．

(2)∵∠B和∠C的外角平分线相交于点P,

∴∠PBC=∠EBC, ∠PCB=∠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°- (∠EBC+∠FCB)=50°