Question

⭕️⭕️⭕️CHALLENGE ⭕️⭕️⭕️





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The sides AB & AC are produced to P&Q.
The bisector of angle PBC and angle QCB intersect at O . Prove that angle BOC =90-1/2of angle A.

Draw figure by ur own .

from triangle chapter .
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Answer 1

Given:

BO is the bisector of ∠PBC and CO is the bisector of ∠QCB.

Proof:

BO is the bisector of ∠PBC.

∴ ∠CBO = (1/2) ∠PBC     ---------------- (i)

=> ∠PBC + y = 180°

=> ∠PBC = 180° - y        ----------------- (ii)

Substitute (ii) in (i), we get

⇒ ∠CBO = (1/2) * (180 - y)

⇒ ∠CBO = 90° - (y/2)

Also,

CO is the bisector of ∠QCB:

∠BCO = (1/2)∠QCB       ---------------- (iii)

=> ∠QCB + z = 180°

⇒ ∠QCB = 180° - z     -------------- (iv)

Substitute (iv) in (iii), we get

∠BCO = (1/2) * (180 - z)

∠BCO = 90° - (z/2)  

Now,

In ΔBOC,

⇒ ∠BOC + ∠BCO + ∠CBO = 180°

⇒ ∠BOC + 90° - (z/2) + 90° - (y/2) = 180°

⇒ ∠BOC = (1/2)[y + z]

⇒ ∠BOC = (1/2)[180 - x]/2

⇒ ∠BOC = 90° - (x°/2)

⇒ ∠BOC = 90° - (A/2)

Hope this helps!

Answer 2

Answer:

plz mark as brainlist answer....