Question

一道几何数学题！！！急急急急！！！快点解答！拜托了！谢谢

（1）如图1，△ABC中，P是∠ABC与∠ACB的角平分线的交点，写出∠BPC与∠A的关系；
（2）如图2，△ABC中，P是∠ABC的角平分线和△ABC的外角∠ACE的角平分线的交点，写出∠BPC与∠A的关系；
（3）如图3，△ABC中，P是外角∠MBC与∠BCN的角平分线，写出∠P与∠A的关系。

Answer 1

(1)∠BPC=∠ABC/2 , ∠PCB=∠ACB/2
∠BPC=180-(∠BPC+∠PCB)=180-(∠ABC/2+∠ACB/2)=180-(180-∠A)/2=90+∠A/2
(2)∠PBC=∠ABC/2 , ∠ACP=∠ACE/2 , ∠ACE=180-∠ACB
∠BPC=180-∠PBC-∠BCP=180-∠ABC/2-(∠ACB+∠ACP)=180-∠ABC/2-[∠ACB+(180-∠ACB)/2]=180-∠ABC/2-(90+∠ACB/2)=90-(∠ABC+∠ACB)/2=90-(180-∠A)/2=∠A/2
(3)∠PBC=∠MBC/2 , ∠MBC=180-∠ABC
∠PCB=∠BCN/2 , ∠BCN=180-∠ACB
∠P=180-(∠PBC+∠PCB)=180-(∠MBC+∠BCN)/2=180-(360-∠ABC-∠ACB)/2=180-360/2+(∠ABC+∠ACB)/2=(180-∠A)/2=90-∠A/2

Answer 2

1. <p=180-<pac-,pca     <p=180-1/2(<b-<c)          <a=180-<b-<c

   <p-<a=180-1/2(<b-<c) -180-<b-<c

             =1/2(<b-<c) 

2.    <a+1/2<b=<p+1/2<ace      <a+1/2<b=<p+1/2(<a+<b)  <a=1/2<a+<p

3.  <p=180-<pbc-<pcn                   

     <p=180-1/2<mbc -1/2<ncb

    <p=180-1/2(<a+<acb)-1/2(<a+<abc)

      <p=180-1/2(<a+<acb+<a+<abc)

     <p=180-1/2(<a+180)

    <p=180-1/2<a-90

   <p=90-1/2<a

                                                                                                    

Answer 3

（1）∠BPC=180-1/2(∠ABC+∠ACB)
∠A=180-(∠ABC+∠ACB)
∠BPC=90+1/2∠A
（2）∠ACE=∠A+∠ABC
∠PCE=∠BPC+∠PBC=1/2∠ACE
∠ABC=2∠PBC
∠BPC=1/2∠A
（3）∠P=180-(∠PBC+∠PCB)
=180-1/2(∠MBC+∠BCN)
=180-1/2(180-∠ABC+180-∠ACB)
=180-1/2(360-(∠ABC+∠ACB))
=180-1/2(360-(180-∠A))
=90-1/2∠A

Answer 4

解：
∵∠A+∠ABC+∠ACB＝180
∴∠ABC+∠ACB＝180-∠A
∵BP平分∠ABC
∴∠PBC＝∠ABC/2
∵CP平分∠ACB
∴∠PCB＝∠ACB/2
∵∠BPC+∠PBC+∠PCB＝180
∴∠BPC＝180-（∠PBC+∠PCB）
＝180-（∠ABC/2+∠ACB/2）
＝180-（∠ABC+∠ACB）/2
＝180-（180-∠A）/2
＝90+∠A/2

Answer 5

解：﹙1﹚，∠p=90°+½∠A
﹙2﹚,∠p=½∠A
﹙3﹚,∠p=90°-½∠A

Answer 6

1,a=180度-2p 因为180-1/2∠ABC-1/2∠ACB=∠P且180-∠ABC-∠ACB=∠A
2,1/2∠ABC+1/2（180度-（180度-∠A-∠ABC））+∠P=180度->∠ABC+1/2∠A+∠P=180度
3，∠ABC+∠A+∠ACB=180度，1/2(180度-∠ABC)+1/2(180度-∠ACB)+∠P=180度->2∠A+∠P=180度