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Question
The bisectors of the equal angles B and C of an isosceles triangle ABC meet at O. Prove that AO bisects angle A.
Solution
In ΔABC, we have AB = AC
⇒ ∠B = ∠C .......[angles opposite to equal sides are equal]
⇒ `[ 1 ] / [ 2 ]` ∠B = `[ 1 ] / [ 2 ]` ∠C
⇒ ∠OBC = ∠OCB ..........(i)
⇒ OB = OC ...........(ii) [angles opposite to equal sides are equal]
Now,
In ΔABO and ΔACO,
AB = AC ......[Given]
∠OBC = ∠OCB ...[From (i)]
OB = OC ...[From (ii)]
ΔABO ≅ ΔACO ...[ SAS criterion ]
⇒ ∠BAO = ∠CAO .......[ c. p . c .t ]
Therefore, AO bisects ∠BAC.
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