In a <span lang="

Given that ABC is a triangle.

BP and CP are internal bisector of B and C respectively

BQ and CQ are external bisector of B and C respectively.

External B = 180o - B

External C = 180o - C

In

BPC + B + C = 180o

BPC = 180o - (B + C) (i)

In

BQC + (180o - B) + (180o - C) = 180o

BQC + 180o - (B + C) = 180o

BPC + BQC = 180o [From (i)]

Hence, proved

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

Prove that the anRS Aggarwal & V Aggarwal - Mathematics

If the sides of aRD Sharma - Mathematics

D is any point onNCERT Mathematics Exemplar

In a triangle RD Sharma - Mathematics

In Δ ABCRD Sharma - Mathematics

In Δ PQRRD Sharma - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

Prove that the peRD Sharma - Mathematics

In Fig. 10.25, <iRD Sharma - Mathematics

In a <span lang="RD Sharma - Mathematics