# In ∆ABC, AB=AC and the bisectors of ∠B and ∠C meet at a point O. prove that BO=CO and the ray AO is the bisector of ∠A. Given: In ∆ABC, AB=AC and the bisectors of B and C meet at a point O.

To prove: BO=CO and BAO = CAO

Proof:

In , ∆ABC we have,

OBC = B

OCB = C

But B = C … given

So, OBC = OCB

Since the base angles are equal, sides are equal

OC = OB …(1)

Since OB and OC are bisectors of angles B and C respectively, we have

ABO = B

ACO = C

∴∠ABO = ACO …(2)

Now in ∆ABO and ∆ACO

AB = AC … given

ABO = ACO … from 2

BO = OC … from 1

Thus by SAS property of congruence,

∆ABO ∆ACO

Hence, we know that, corresponding parts of the congruent triangles are equal

BAO = CAO

ie. AO bisects A

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