In triangle ABC, AB =AC and the bisector of angle B and angle C meet at point O.Prove that BO=CO and the ray AO is the bisector of angle A
Answers
Given : In triangle ABC, AB =AC and the bisector of angle B and angle C meet at point O.
To Find : Prove that BO=CO and the ray AO is the bisector of angle A
Solution:
In triangle ABC, AB =AC
=> ∠B = ∠C ( angles opposite to equal sides in a triangle are equal )
bisector of angle B and angle C meet at point O.
=> ∠ABO = ∠CBO = ∠B/2
∠ACO = ∠BCO = ∠C/2
∠B = ∠C => ∠B/2 = ∠C/2 =>
∠CBO = ∠BCO = ∠ABO = ∠ACO
in triangle BOC
∠CBO = ∠BCO
=> BO = CO
in Δ AOB and ΔAOC
AO = AO common
BO = CO Shown above
AB = AC given
=> Δ AOB ≅ ΔAOC
=> ∠AOB = ∠AOC
∠AOB + ∠AOC = ∠A
=> ∠AOB = ∠AOC = ∠A/2
Hence AO is the bisector of angle A
QED
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Step-by-step explanation:
It is given that AB = AC and the bisectors ∠ B and ∠ C meet at a point O Consider △ BOC So we get ∠ BOC = ½ ∠ B and ∠ OCB = ½ ∠ C It is given that AB = AC so we get ∠ B = ∠ C So we get ∠ OBC = ∠ OCB We know that if the base angles are equal even the sides are equal So we get OB = OC ……. (1) ∠ B and ∠ C has the bisectors OB and OC so we get ∠ ABO = ½ ∠ B and ∠ ACO = ½ ∠ C So we get ∠ ABO = ∠ ACO …….. (2) Considering △ ABO and △ ACO and equation (1) and (2) It is given that AB = AC By SAS congruence criterion △ ABO ≅ △ ACO ∠ BAO = ∠ CAO (c. p. c. t) Therefore, it is proved that BO = CO and the ray AO is the bisector of ∠ A.Read more on Sarthaks.com - https://www.sarthaks.com/718710/in-abc-ac-and-the-bisectors-and-meet-at-point-prove-that-bo-co-and-the-ray-ao-is-the-bisector-of
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