Question

如图,在△ABC中,BD,CE分别是∠ABC,∠ACB的平分线.

Answer 1

∠BPC=∠ECA+∠BDC,∠BDC=∠A+∠ABD,∵BD,CE分别是∠ABC,∠ACB的平分线,∴∠BPC=∠A+1/2∠ABC+1/2∠ACB,（1）又∵∠ABC=60°,∠ACB=40°,∴∠A=80°,∴∠BPC=80°+30°+20°=130°.（2）∵∠A=80°,∴∠ABC+∠ACB=100°,∴∠BPC=∠A+1/2∠ABC+1/2∠ACB=80°+1/2（∠ABC+∠ACB）=130°.（3）若∠A=X°,则∠BPC=∠A+1/2∠ABC+1/2∠ACB=X°+1/2（∠ABC+∠ACB）=X°+1/2（180°-∠A）=X°+90°-1/2X°∴∠BPC=1/2X°+90°