Question

The sides AB and AC of a ∆ ABC are produced to P and Q respectively. If the bisectors of ∠PBC and ∠QCB intersect at O, then

$\rm \angle \: BOC = 90 \degree - \frac{1}{2} \angle A$
​

Answer 1

Since ∠ABC and ∠CBP form a linear pair.

∠ABC+∠CBP=180°

∠B+2∠1=180°

[BO is the bisector of ∠ CBP]

2∠1=180° −∠B

$∠1=90°− \frac{1}{2} ∠B \: \: \: - - - - (1)$

Again, ∠ACB and ∠QCB form a linear pair.

Therefore,

∠ACB+∠QCB=180°

∠C+2∠2=180°

2∠2=180°−∠C

$∠2=90 \degree− \frac{1}{2} ∠C      - - - (2)$

In △BOC, we have

∠1+∠2+∠BOC=180°

$\small{90− \frac{1}{2} ∠B+90 \degree - \frac{1}{2} ∠C+∠BOC=180 \degree}$

$\small{180 \degree - \frac{1}{2} (∠B+∠C)+∠BOC=180 \degree}$

$∠BOC= \frac{1}{2} (∠B+∠C)$

$∠BOC= \frac{1}{2} (180 \degree - \angle \: A )$

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

Answer 2

Step-by-step explanation:

Given:-

A ∆ ABC in which AB and AC are produced to P and Q respectively. The bisectors of ∠PBC and ∠QCB intersect at O.

To Prove:-

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

Proof:-

Since ∠ABC and ∠CBP from a linear pair.

∠ABC + ∠CBP = 180°

→ ∠B + 2∠1 = 180° [BO is the bisector of ∠CBP]

→ 2∠1 = 180° - ∠B

→ ∠1 = 90- ½ ∠B -------(i)

Again, ∠ACB and ∠QCB form a linear pair

∠ACB + ∠QCB = 180°

→ ∠C + 2∠2 = 180° [DC is the bisector of ∠QCB]

→ 2∠2 = 180° - ∠C

→∠2 = 90° - ½ ∠C --------(ii)

Also,

∠A + ∠B + ∠C = 180°

→ ∠B + ∠C = 180° - ∠A

In ∆BOC, we have

∠1 + ∠2 + ∠BOC = 180°

→ 90° - ½∠B + 90° -½∠C + ∠BOC = 180°[Using(i) & (ii)]

→ ∠BOC = ½(∠B + ∠C)

→ ∠BOC = ½(180° - ∠A) [ Proved above]

→ ∠BOC = 90° - ½∠A

Hence proved