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Question
In ΔABC, AB = AC and the bisectors of angles B and C intersect at point O.
Prove that : (i) BO = CO
(ii) AO bisects angle BAC.
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Solution
In ΔABC,
AB = AC
⇒ ∠B = ∠C ...( angles opposite to equal sides are equal )
⇒ `1/2 ∠"B" = 1/2∠"C"`
⇒ ∠OBC = ∠OCB ...[ ∵ OB and OC are bisectors of ∠B and ∠C respectively, ∠OBC = `1/2∠"B" and ∠"OCB" = 1/2∠"C"` ] ...(i)
⇒ OB = OC ...( Sides opposite to equal angles are equal ) ...(ii)
Now, in ΔABO and ΔACO,
AB = AC ...( given )
∠OBC = ∠OCB ...[ from(i) ]
OB = OC ...[ from(ii) ] ...( proved )
∴ ΔABO ≅ ΔACO ...( by SAS congruence criterion )
⇒ ∠BAO = ∠CAO ...( c.p.c.t. )
⇒ AO bisects ∠BAC ...(proved)
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