Question

The bisectors of ∠B and ∠C of an isosceles ΔABC with AB = AC intersect each other at a point O. Show that the exterior angle adjacent to ∠ABC is equal to ∠BOC.

Answer 1

Given: In an isosceles ΔABC, 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 ΔABC,

$\because$ AB = AC (Given)
$\therefore$ ∠ACB = ∠ABC (Angle opposite to equal sides are equal)

$$\begin{gathered} \Rightarrow \frac{1}{2}\angle ACB=\frac{1}{2}\angle ABC \\ \Rightarrow \angle OCB=\angle OBC.....\left(\mathrm{i}\right) \\ \left(\mathrm{Given},BO\mathrm{and}CO\mathrm{are}\mathrm{angle}\mathrm{bisector}\mathrm{of}\angle ABC\mathrm{and}\angle ACB,\mathrm{respectively}\right) \end{gathered}$$

In ΔBOC,

$$\begin{gathered} \angle OBC+\angle OCB+\angle BOC=180°\left(\mathrm{By}\mathrm{angle}\mathrm{sum}\mathrm{property}\mathrm{of}\mathrm{triangle}\right) \\ \Rightarrow \angle OBC+\angle OBC+\angle BOC=180°\left[\mathrm{From}\left(\mathrm{i}\right)\right] \\ \Rightarrow 2\angle OBC+\angle BOC=180° \\ \Rightarrow \angle ABC+\angle BOC=180°\left(BO\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{bisector}\mathrm{of}\angle ABC\right).....\left(\mathrm{ii}\right) \end{gathered}$$

Also, DBC is a straight line.
So, $\angle ABC+\angle DBA=180°\left(\mathrm{Linear}\mathrm{pair}\right).....\left(\mathrm{iii}\right)$

From (ii) and (iii), we get
$$\begin{gathered} \angle ABC+\angle BOC=\angle ABC+\angle DBA \\ \therefore \angle BOC=\angle DBA \end{gathered}$$