Question

七年级下学期题目求解 急！！！！！！

在△ABC中，若点P是∠ABC和∠ACB的角平分线的交点。若BP、CP均变为∠ABC、∠ACB的邻补角∠CBF、∠BCE的角平分线，如图（1 )(2)(3)问∠A与∠P的关系？三个图都要几何公式解哦

Answer 1

1)∠PBC+∠PCB=1/2(∠ACB+∠ABC)=1/2(180-∠A)=90-1/2∠A
所以∠BPC=180-(90-1/2∠A)=90+1/2∠A
2.)∠ABP+∠ACP=1/2(180-∠B)+1/2(180-∠C)=180-1/2∠B-1/2∠C
=180-1/2(180-∠A)=90+1/2∠A
所以∠BPC=90-1/2∠A
3.）角BPC=角PCD-角PBD
=1/2(x+ABD)-1/2ABD
=1/2∠A