Question

In TriangleABC, internal bisector of Angle B and angle C meet at P and external bisector of Angle B and angle C meet at Q prove that angle B P C + Angle B Q C is equal to 180 degree

Answer 1

Answer:

From the figure, it shows that

∠ACB and ∠QCB form a linear pair,

So, ∠ACB + ∠QCB = 180º

=> ∠ACB/2 + ∠QCB/2 = 180º /2

=> ∠ACB/2 + ∠QCB/2 = 90º

Again since PC and QC are the angle bisectors

=> ∠PCB + ∠BCQ = 90º

=> ∠PCQ = 90º

Similarlry,

∠PBQ = 90º

In qradrilateral BPCQ,

Sum of all four angles = 360º

=> ∠BPC + ∠PCQ + ∠CQB + ∠QBP = 360º

=> ∠BPC + ∠BQC + 180º = 360º

=> ∠BPC + ∠BQC = 360º - 180º

=> ∠BPC + ∠BQC = 180º

=> ∠P = ∠Q = 90º [Since ∠PCQ = ∠PBQ = 90º]