Question

如图,已知,三角形ABC中,∠A=α,BP平分∠ABC,CP平分∠ACD,求P的度数

Answer 1

在△BCP中,

∵∠PBC+∠P+∠PCB=180°

∴∠P=180°-1/2∠ABC-（∠PCA+∠ACB)

=180°-1/2∠ABC-(1/2∠ACD+∠ACB)

=180°-1/2∠ABC-[1/2(∠A+∠ABC)+∠ACB)

=180°-1/2∠ABC-1/2∠A-1/2∠ABC-∠ACB

=180°-(∠ABC+∠ACB)-1/2∠A

=∠A-1/2∠A

=1/2∠A

=(1/2)α

因为看不到图,只能凭着题目和自己的理解写,过程写得很细,你再看一下吧,应该是对的,希望能帮到你^^