Bisectors of the angles B and C of an isosceles triangle ABC with AB = AC intersect each other at O. Show that external angle adjacent to ∠ ABC is equal to ∠ BOC.

Question

Philoid · Accepted Answer

Given: ΔABC with AB = AC

In ΔABC,

AB = AC (given)

⇒ ∠ACB = ∠ABC (opposite angles to equal sides are equal)

1/2 ∠ACB = 1/2 ∠ABC (divide both sides by 2)

⇒ ∠OCB = ∠OBC …(1) (As OB and OC are bisector of ∠B and ∠C)

Now, in ΔBOC,

∠OBC + ∠OCB + ∠BOC = 180° (angle sum property)

∠OBC + ∠OBC + ∠BOC = 180° (from (1))

⇒ 2 ∠OBC + ∠BOC = 180°

⇒ ∠ABC + ∠BOC = 180° (because OB is bisector of ∠B)

⇒ 180° - ∠DBA + ∠BOC = 180°

⇒ - ∠DBA + ∠BOC = 0

⇒∠DBA = ∠BOC Hence proved.

Bisectors of the angles B and C of an isosceles triangle ABC with AB = AC intersect each other at O. Show that external angle adjacent to ∠ABC is equal to ∠BOC.