Data: In an isosceles triangle ABC, with AB = AC, the bisectorts of ∠B and ∠C intersect each other at O. Join A to O.
To Prove:
(i) OB = OC
(ii) AO bisects ∠A.
Proof:
(i) In ∆ABC,
AB = AC
∴ ∠ABC = ∠ACB
\(\frac{1}{2}\)ABC = \(\frac{1}{2}\)ACB
∠OBC = ∠OCB.
In ∆OBC
Now, ∠OBC = ∠OCB is proved.
∴ ∆OBC is an isosceles triangle.
∴ OB = OC.
(ii) In ∆AOB and ∆AOC,
AB = AC (Data)
OB = OC (proved)
AO is common.
Side, Side, Side postulate.
∴ ∆AOB ≅ ∆AOC
∴ ∠OAB = ∠OAC
∴ AO bisects ∠A.