8. In a triangle ABC, the internal bisectors of angle B and angle C meet at P and the external bisectors of ZB and ZC meet at Q, prove that angle BPC + angle BQC = 180º.
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In a triangle ABC, the internal bisectors of angle B and C meet at P and the external bisector of the angle B and C meet at Q. Prove that : ∠BPC + ∠BQC = 2 rt.
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Step-by-step explanation:
∠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]
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