Question

如图 三角形abc的两条内角平分线BD,CE相交于点P,两条外角平分线BQ,CQ相交于点Q.

Answer 1

http://hi.baidu.com/yzwjmx%5F/album/item/67aec0a876540dde1f17a2e7.html

∵BD平分∠ABC,

∴∠DBC＝∠ABC/2

同理∠DCB＝∠ACB/2

∵∠DBC＋∠DCB＋∠BPC＝180°

∴∠BPC＝180°－∠DBC－∠DCB

＝180－(∠ABC＋∠ACB)/2

又∵∠A＋∠ABC＋∠ACB＝180°

∴∠ABC＋∠ACB＝180°－∠A

∴∠BPC＝180°－(180°－∠A)/2

即∠BPC＝90°＋∠A/2

∠A=50° ,∠BPC＝90°＋25°=115°