# In the second semester of grade seven, two outer Angle bisectors were used to find angles and models

In the second semester of grade 7, we need to skillfully use a variety of triangle models, such as the Angle problem of bisectors of two outer angles introduced in this article.As shown in the figure, in △ABC, the bisectors of the outer angles of ∠ABC and ACB intersect at point 0. Let ∠A=m, and find the measure of ∠BOC.To find the relationship between two outer Angle bisectors in a triangle and another Angle, we can use the triangle inner Angle sum theorem, the nature of the outer Angle, the nature of the Angle bisector, Angle and difference relations to solve the problem, get two angles when love you.According to this result, we can solve for the Angle between the bisectors of two outer angles.As shown in figure 1, bisectors of two exterior angles of △ABC intersect at point P. If Angle A=40°, find the measure of Angle BPC.Analysis: according to the definition of Angle bisector that an outer Angle of a triangle is equal to the sum of two inner angles which are not adjacent to it, Angle PBC and Angle PCB can be expressed, and then according to the triangle inner Angle sum theorem, the solution can be obtained.Solution:∵, < < CBD BCE bisector intersect at point P, ∴ < PBC = 1/2 (< A + < ACB), < PCB = 1/2 (ABC) A + < <, ∴ < = 1/2 PBC + < PCB (< A + < ACB ABC + + < < A), A + ∵ < < ACB ABC + < = 180 °,∴ < PBC + = 90 ° + 1/2 < < PCB A, in the delta of PBC, < the BPC = 180 ° - (< PBC + < PCB) = 180 ° to 90 ° (+ 1/2 < A) < = 90 ° - 1/2 A, ∵ < A = 40 °, ∴ < BPC = 40 ° 90 ° - 12 x = 90 ° ~ 20 ° 70 °.In the pentagon ABCDE, the external angles of ∠BCD and EDC are FCD and GDC, respectively. CP and DP bisect FCD and GDC respectively and intersect at point P. If ∠A=140°, ∠B=120°, and ∠E=90°, find the measure of Angle P.(n-2) •180°, we can get the sum of Angle BCD and Angle EDC, so as to get the sum of two adjacent outer angles, and then according to Angle bisector and triangle inner Angle sum theorem can get the answer.Solution: Sum of interior angles of polygons(n - 2), 180 ° = 540 °, ∴ < BCD + < EDC = 540 ° to 140 ° to 120 ° to 90 ° = 190 °, and ∵ CP and DP were < BCD, < EDC exterior Angle bisector, ∴ < PCD + < PDC = 1/2 (360 ° - < BCD - < EDC) = 85 °,Angle CPD=180°-85°=95°