Question

in triangle ABC, the sides AB and AC are produced to P and Q respectively. the bisectors of angle QCB intersect at a point O. prove that angle BOC=90°-1/2 angle A​

Answer 1

Please refers to the attachment o know which is

∠x, ∠y, and ∠z

Solution : Ray BO is the bisectors of ∠CBE

Therefore,

∠CBO = 1/2 ∠CBE

= 1/2( 180° - y)

= 90° - y/2. .............( i )

Similarly, ray CO is the bisector of ∠ BCD

Therefore, ∠ BOC = 1/2 ∠ BCD

= 1/2 (180° - z)

= 90° - z/2 ..........( ii )

IN ∆ BOC, ∠BOC + ∠BCO + ∠CBO = 180°.......( iii )

Substituting ( i ) and ( ii ) in ( iii ), we get

∠BOC + 90° - z/2 + 90° - y/2 ÷ 180°

so, ∠BOC = z/2 + y/2

=> ∠BOC = 1/2 ( y + z )........( iii )

but, x + y + z = 180° ( sum of interior angle of triangle is 180°)

Therefore,

y + z = 180° - x

Therefore eq (iv) becomes

∠BOC = 1/2 (180° - x)

= 90° - x/2

=90° - 1/2 ∠BAC

HENCE, proved

Answer 2

Answer:

Proved!

Step-by-step explanation:

Correct Question:

The sides AB and AC of triangle ABC are produced to P and Q respectively. The bisectors of exterior angles at B and C of triangle ABC meet at O.  Prove that ∠BOC = 90° -1/2 ∠A

Solution:

To Prove that:

BOC = 90° -1/2 ∠A

-------------------------

Given that,

In triangle ABC,

(∠ABC + ∠ACB) = 180 - ∠A

Now,  

∠CBP = ∠A +∠ACB

So,

∠CBO = 1/2 ∠ A + 1/2 ∠ACB

Similarly,

∠BCO = 1/2 ∠A + 1/2 ∠ABC

In triangle BOC,

∠BOC = 180° - ∠CBO - ∠BCO

∠BOC = 180° - (∠CBO + ∠BCO)

∠BOC = 180° - (1/2 ∠A + 1/2∠ABC+1/2 ∠ A + 1/2 ∠ACB)

∠ BOC = 180° - (∠A + 1/2∠ABC + 1/2 ∠ACB)

∠BOC = 180° - (∠A + 90 - 1/2 ∠A)

∠BOC = 90° - 1/2 ∠A                          (Proved)

Hence, Proved!