图20题
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/06/25 11:31:55
![](http://img.wesiedu.com/upload/2/0f/20fba9f12cb07538f724bf89cfd51ab3.jpg)
![图20题](/uploads/image/z/19053397-37-7.jpg?t=%E5%9B%BE20%E9%A2%98)
解题思路: ∵∠ABC的平分线BF与△ACB的外角∠ACE的平分线CD相交于点D, ∴∠PCE=∠ACE,∠PBC=∠ABC, ∵∠PCE是△BCP的外角, ∴∠P=∠PCE-∠PBC =1/2∠ACE-1/2∠ABC =1/2(∠A+∠ABC)-1/2∠ABC =1/2∠A+1/2∠ABC-1/2∠ABC =1/2∠A=1/2×50° =25°
解题过程:
解:(1) 20°
(2)30°
(3) ∵∠ABC的平分线BF与△ACB的外角∠ACE的平分线CD相交于点D,
∴∠PCE=∠ACE,∠PBC=∠ABC,
∵∠PCE是△BCP的外角,
∴∠BPC=∠PCE-∠PBC
=1/2∠ACE-1/2∠ABC
=1/2(∠A+∠ABC)-1/2∠ABC
=1/2∠A+1/2∠ABC-1/2∠ABC
=1/2∠A
解题过程:
解:(1) 20°
(2)30°
(3) ∵∠ABC的平分线BF与△ACB的外角∠ACE的平分线CD相交于点D,
∴∠PCE=∠ACE,∠PBC=∠ABC,
∵∠PCE是△BCP的外角,
∴∠BPC=∠PCE-∠PBC
=1/2∠ACE-1/2∠ABC
=1/2(∠A+∠ABC)-1/2∠ABC
=1/2∠A+1/2∠ABC-1/2∠ABC
=1/2∠A