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