Question

If internal bisector of angle B and angle C intersect at P, prove that angle PBO=90 and angle BOC+angle BPC=180

Answer 1

Answer:

∠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]

Step-by-step explanation:

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Answer 2

Answer:

ok

Step-by-step explanation:

Answer:

∠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]

Step-by-step explanation:

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