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Question
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:
- OB = OC
- AO bisects ∠A
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Solution
(i) ABC is an isosceles triangle in which AB = AC
∠C = ∠B ...[Angles opposite to equal sides in a triangle are equal.]
⇒ ∠OCA + ∠OCB = ∠OBA + ∠OBC
⇒ ∠OCB + ∠OCB = ∠OBC + ∠OBC
∵ OB bisects ∠B.
∴ ∠OBA = ∠OBC
And OC bisects ∠C.
∴ ∠OCA = ∠OCB
⇒ 2∠OCB = 2∠OBC
⇒ ∠OCB = ∠OBC
Now, in △OBC,
∠OCB = ∠OBC ...[Proved above]
∴ OB = OC ...[Sides opposite to equal angles]
(ii) Now, in △AOB and △AOC,
AB = AC ...[Given]
∠OBA = ∠OCA
∠B = ∠C
BO bisects ∠B and CO bisects ∠C.
∠OBA = ∠OCA
OB = OC
∴ △AOB ≌ △AOC ...[By SAS congruence rule]
⇒ ∠OAB = ∠OAC ...[Corresponding parts of congruent triangles]
So, AO bisects ∠A.
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