Question

$\tt \dag \red{Q} \green{u} \blue{e} \purple{s} \red{t} \green{i} \blue{o} \purple{n}$
In the adjoining figure, A ∆ABC in which AB and AC are produced to P and Q respectively. The bisectors of ∠PBC and ∠QCB intersect at O.

$\tt \: Prove : \: \angle \: BOC = 90 \degree - \frac{1}{2} \angle \: A$

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Answer 1

In a ABC∆, the sides AB and AC are produced to P and Q respectively. The bisectors of ∠ PBC and ∠ QCB intersect at a point O. Prove that ∠ BOC =90 ° − 1/2 ∠A

ok let's understand the concept

Given :

• bisectors of angle PBC and angle QCB intersect at O.

 To find :

• prove that ∠BOC = 90° -  (1/2) (∠A)  

Solution:

• ∠BAC = ∠A

Exterior angle = Sum of opposite interior angles

=>  ∠ PBC = ∠C + ∠A

=> ∠ QCB = ∠B + ∠A

=> ∠CBO = (1/2)∠PBC  = (1/2) (∠C + ∠A)

=> ∠BCO = (1/2)∠QCB   = (1/2) (∠B + ∠A)

 ∠CBO  + ∠BCO  + ∠BOC = 180°   Sum of angles 0 a triangle )

=>  (1/2) (∠C + ∠A) +  (1/2) (∠B + ∠A)  +  ∠BOC = 180°

=> (1/2) (∠C + ∠A + ∠B) +  (1/2) (∠A)  +  ∠BOC = 180°

=> (1/2) (180°) +  (1/2) (∠A)  +  ∠BOC = 180°

=> 90°  +  (1/2) (∠A)  +  ∠BOC = 180°

=> ∠BOC = 90° -  (1/2) (∠A)  

$\mathfrak{Hence \: Proved}$

$\sf{_________}$

Answer 2

EXPLANATION.

In the adjoining figure,

ΔABC in which AB and AC are produced to P and Q.

Bisectors of ∠PBC and ∠QCB intersects at O.

As we know that,

To prove : ∠BOC = 90° - 1/2∠A.

BO is bisector of ∠CBP.

⇒ ∠CBO = ∠PBO = 1/2∠CBP.

CO is bisector of ∠BCQ.

⇒ ∠BCO = ∠QCO = 1/2∠BCQ.

∠CBP is an exterior angles of ΔABC.

⇒ ∠CBP = ∠A + ∠2.

⇒ 1/2∠CBP = 1/2(∠A + ∠2).

⇒ ∠CBO = 1/2(∠A + ∠2).

∠BCQ is an exterior angles of ΔABC.

⇒ ∠BCQ = ∠A + ∠1.

⇒ 1/2∠BCQ = 1/2(∠A + ∠1).

⇒ ∠BCO = 1/2(∠A + ∠1).

As we know that,

Sum of three angles of a triangle is = 180°.

⇒ ∠A + ∠1 + ∠2 = 180°.

In ΔOBC.

⇒ ∠BOC + ∠BCO + ∠CBO = 180°.

⇒ ∠BOC + 1/2(∠A + ∠1) + 1/2(∠A + ∠2) = 180°.

⇒ ∠BOC + 1/2(∠A + ∠A + ∠1 + ∠2) = 180°.

⇒ ∠BOC + 1/2(∠A + 180°) = 180°.

⇒ ∠BOC + ∠A/2 + 180°/2 = 180°.

⇒ ∠BOC + ∠A/2 + 90° = 180°.

⇒ ∠BOC = 180° - 90° - ∠A/2.

⇒ ∠BOC = 90° - ∠A/2.

Hence proved.