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Question
Bisectors of the angles B and C of an isosceles triangle ABC with AB = AC intersect each other at O. Show that external angle adjacent to ∠ABC is equal to ∠BOC
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Solution
Given: ΔABC is an isosceles triangle in which AB = AC, BO and CO are the bisectors of ∠ABC and ∠ACB respectively intersect at O.
To show: ∠DBA = ∠BOC
Construction: Line CB produced to D.
Proof: In ΔABC, AB = AC ...[Given]
∠ACB = ∠ABC ...[Angles opposite to equal sides are equal]
⇒ `1/2 ∠ACB = 1/2 ∠ABC` ...[On dividing both sides by 2]
⇒ ∠OCB = ∠OBC ...(i) [∵ BO and CO are the bisectors of ∠ABC and ∠ACB]
In ΔBOC, ∠OBC + ∠OCB + ∠BOC = 180° ...[By angle sum property of a triangle]
⇒ ∠OBC + ∠OBC + ∠BOC = 180° ...[From equation (i)]
⇒ 2∠OBC + ∠BOC = 180°
⇒ ∠ABC + ∠BOC = 180° ...[∵ BO is the bisector of ∠ABC]
⇒ 180° – ∠DBA + ∠BOC = 180° ...[∵ DBC is a straight line]
⇒ – ∠DBA + ∠BOC = 0
⇒ ∠DBA = ∠BOC
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