$I n Δ ABC, AB = AC and the bisectors of ∠ B a n d ∠ C meet at a point O. Prove that BO = CO and the ray AO is the bisector of ∠ A .$

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Solution

In triangle ABC, we have:

AB = AC (Given)

⇒∠B=∠C

⇒ $\frac{1}{2}$ ∠B= $\frac{1}{2}$ ∠C

⇒∠OBC=∠OCB

⇒BO=CO

Now, in △AOB and △AOC, we have: OB=OC (Proved)

AB=AC (Given)

AO=AO (Common)

i.e., ∠BAO=∠CAO (Corresponding angles of congruent triangles)

∴△AOB≅△AOC (SSS criterion)

So, it shows that ray AO is the bisector of ∠A.

Hence, proved.

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$I n Δ ABC, AB = AC and the bisectors of ∠ B a n d ∠ C meet at a point O. Prove that BO = CO and the ray AO is the bisector of ∠ A .$