the sides AB and AC of a triangleABC are produced to p and q respectively if the bisectors of angle PBC and angle QCB intersect at O then prove that angle BOC 90°-1÷2 angle A
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Answer:
Since ∠ABC and ∠CBP form a linear pair.
∠ABC+∠CBP=180
o
∠B+2∠1=180
o
[BO is the bisector of ∠ CBP]
2∠1=180
o
−∠B
∠1=90
o
−
2
1
∠B .....(1)
Again, ∠ACB and ∠QCB form a linear pair.
Therefore,
∠ACB+∠QCB=180
o
∠C+2∠2=180
o
2∠2=180
o
−∠C
∠2=90
o
−
2
1
∠C ...(2)
In △BOC, we have
∠1+∠2+∠BOC=180
o
90−
2
1
∠B+90
o
−
2
1
∠C+∠BOC=180
o
180
o
−
2
1
(∠B+∠C)+∠BOC=180
o
∠BOC=
2
1
(∠B+∠C)
∠BOC=
2
1
(180
o
−∠A)
∠BOC=90
o
−
2
1
∠A
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