Question

如图所示，已知△ABC中，∠ABC和∠ACB的外角平分线相交于点P.

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°