Answer :
Given: ABC is a triangle . Let AB be produced to P and AC be produced to Q. Let the bisectors of ∠ PBC and ∠ QCB meet at O.
To prove: ∠ BOC = 90° -
Proof: In Δ ABC,
∠ A + ∠ B + ∠ C = 180° (Sum of the three angles of a triangle is 180° )
∠ A +
∠ B +
∠ C =
× 180°
∴
= 90° -
……………………….(1)
= 90° -
In Δ BOC,
∠ BOC + ∠ CBO + ∠ BCO = 180°
∠ BOC = 180° - (∠ CBO + ∠ BCO) ………………(2)
∠ PBC = 180° - ∠ ABC (Since ∠ ABC and ∠ CBP form a linear pair)
Dividing by 2,
∠ PBC =
(180° - ∠ ABC)
∴ ∠ CBO = 90°
…………………..(3)
Ð QCB = 180° - ∠ ACB (Since ∠ QCB and ∠ ACB form a linear pair)
∠ QCB =
(180° - Ð ACB)
∴ ∠ BCO = 90° -
………………………(4)
Substituting (3) and (4) in (2),
∠ BOC = 180° - (∠ CBO + ∠ BCO)
= 180° - (90° -
+ 90° -
)
= 180° - (180° -
-
)
=
+
…………………………….(5)
But from (1),
+
= 90° -
Substituting in (5),
∠ BOC = 90° -
To prove: ∠ BOC = 90° -

Proof: In Δ ABC,
∠ A + ∠ B + ∠ C = 180° (Sum of the three angles of a triangle is 180° )




∴




In Δ BOC,
∠ BOC + ∠ CBO + ∠ BCO = 180°
∠ BOC = 180° - (∠ CBO + ∠ BCO) ………………(2)
∠ PBC = 180° - ∠ ABC (Since ∠ ABC and ∠ CBP form a linear pair)
Dividing by 2,


∴ ∠ CBO = 90°

Ð QCB = 180° - ∠ ACB (Since ∠ QCB and ∠ ACB form a linear pair)


∴ ∠ BCO = 90° -

Substituting (3) and (4) in (2),
∠ BOC = 180° - (∠ CBO + ∠ BCO)
= 180° - (90° -


= 180° - (180° -


=


But from (1),



Substituting in (5),
∠ BOC = 90° -

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