Question

In triangle ABC angle cbd and angle bce an two exterior angles the bisector of the exterior angle meet of o proved that angle bic = 90°1/2 angle A​

Answer 1

Answer:

In △BOC,

∠BOC+∠OBC+∠OCB=180 (OB and OC bisect ∠B and ∠C respectively)

∠BOC+

2

1

∠B+

2

1

∠C=180

∠BOC=180−

2

1

(∠B+∠C)

∠BOC=180−

2

1

(180−∠A)

∠BOC=180−

2

1

(180−70)

∠BOC=180−90+35

∠BOC=125

∘

Answer 2

Answer:

this is your answer

Step-by-step explanation:

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