Question

26. In given figure the side AB and AC of AABC are produced to point E and D respectively. If
bisector BO and Co of ZCBE and ZBCD respectively meet at point o, then prove that
1
ZBOC = 90°-=ZBAC ..
2
O

Answer 1

Correct question :

• In given figure the side AB and AC of ΔABC are produced to point E and D respectively. If bisector BO and Co of ∠CBE and ∠BCD respectively meet at point o, then prove that ∠BOC = 90°- ½∠BAC ..

Given:

• In ∠ABC, BO and CO are angle bisectors of ZEBC and ZBCD respectively.

To prove :

• ∠BOC = 90 - ½∠BAC

Proof :

We know, exterior angle is equal to the sum of interior Opposite angles.

∠EBC = ∠A + ∠ACB

and ∠BCD = ∠A + ∠ABC

Adding the above equations,

∠EBC + ∠BCD = ∠A +∠A + ∠ACB + ∠ABC

∠EBC + ∠BCD = ∠A + 180°

∠EBC/2 + ∠BCD/2 = ∠A/2 + 180°/2

[Dividing both sides by 2]

∠OBC+ ∠OCB = ½∠A +90°.......( i )

In ∠BOC,

∠BOC + ∠0BC + ∠OCB = 180° [Angle sum property of a triangle]

∠BOC + 90° + ½ ∠A = 180° [Using (I) ]

∠BOC = 180°-90°- ½∠A

∠BOC = 90° - ½ ∠A ...........Hence Proved.

Answer 2

REFER TO THE ATTACHMENT

HOPE IT HELPS :)