Question

$\text{The bisectors of}\angle Band\angle C\text{of an isosceles}\Delta \text{ABC with AB = AC intersect each other at a point O. Show that the exterior angle adjacent to}\angle ABC\text{is equal to}\angle BOC.$

Answer 1

Given: In an isosceles Δ ABC, AB = AC , BO and CO are the bisectors of ∠ ABC and ∠ ACB, respectively.
To prove: ∠ ABD = ∠ BOC
Construction: Produce CB to point D.
Proof:
In Δ ABC,
∵ AB = AC
∴ ∠ ACB = ∠ ABC (Given) (Angle opposite to equal sides are equal)

⇒ $\frac{1}{2}$∠ACB = $\frac{1}{2}$∠ABC

⇒ ∠OCB = ∠OBC .....i

Given, BO and CO are angle bisector of ∠ABC and ∠ACB, respectively

In Δ BOC,

∠OBC+∠OCB+∠BOC=180°

By angle sum property of the triangle

⇒∠OBC+∠OBC+∠BOC=180°

⇒∠ABC+∠BOC=180° .....ii

BO is the angle bisector of ∠ABC

Also, DBC is a straight line.

So, ∠ABC+∠DBA=180°
Linear pair
From i

⇒2∠OBC+∠BOC=180°.....iii

From (ii) and (iii), we get

∠ABC+∠BOC = ∠ABC+∠DBA

∴ ∠BOC = ∠DBA