Question

Question is in the above picture

Answer 1

In ΔABC, AB = AC
⇒ ∠B = ∠C� [Angles opposite to equal sides are equal]
Also OA and OB are bisectors of angles B and C.

⇒ ∠OBC = ∠OCB
∴ OB = OC [Sides opposite to equal angles are equal]
Now consider, Δ’s AOB and AOC
OA = OA (Common side)
AB = AC (Given)
OB = OC (Proved)
ΔAOB ≅ ΔAOC [By SSS congruence criterion]
⇒ ∠OAB = ∠OAC
That is OA is bisector of angle A


In ΔABC, AB = AC
⇒ ∠B = ∠C  [Angles opposite to equal sides are equal]
Also OA and OB are bisectors of angles B and C.

⇒ ∠OBC = ∠OCB
∴ OB = OC [Sides opposite to equal angles are equal]
Now consider, Δ’s AOB and AOC
OA = OA (Common side)
AB = AC (Given)
OB = OC (Proved)
ΔAOB ≅ ΔAOC [By SSS congruence criterion]
⇒ ∠OAB = ∠OAC
That is OA is bisector ∠A.

Answer 2

after this picture
u have to write::::::
Since AB=AC
∠ACB=∠ABC ((angles opp. to equal sides are equal))
1/2 ∠ACB = 1/2 ∠ABC
∠OCB=∠OBC ((from 2 and 3))
hence,,,,
OB=OC ((sides opp. to equal angles r equal))
HENCE PROVED



hope it will help u...