Question

The sides AB and AC of ∆ABC have been produced to D and E, respectively. The bisectors of ∠CBD and ∠BCE meet at O. If ∠A = 40°, find ∠BOC.

Answer 1

Let
$$\begin{gathered} \angle OBC=\angle OBD=\angle 1\left[\because OB\text{is the angle bisector}\right]\text{and} \\ \angle OCB=\angle OCE=\angle 2\left[\because OC\text{is the angle bisector}\right] \end{gathered}$$
Then,
$$\begin{gathered} \angle B+\angle CBD=180°\left[\text{linear pair}\right] \\ \Rightarrow \frac{1}{2}\angle B+\angle 1=90° \\ \Rightarrow \angle 1=90°-\frac{1}{2}\angle B...\left(i\right) \end{gathered}$$
Again,
$$\begin{gathered} \angle C+\angle BCE=180°\left[\text{linear pair}\right] \\ \Rightarrow \frac{1}{2}\angle C+\frac{1}{2}\angle BCE=90° \\ \Rightarrow \frac{1}{2}\angle C+\angle 2=90° \\ \Rightarrow \angle 2=90°-\frac{1}{2}\angle C...\left(ii\right) \end{gathered}$$
Now, in $∆OBC$, we have:
$$\begin{gathered} \angle 1+\angle 2+\angle BOC=180°\left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 90°-\frac{1}{2}\angle B+90°-\frac{1}{2}\angle C+\angle BOC=180° \\ \Rightarrow \angle BOC=\frac{1}{2}\angle B+\angle C \\ \Rightarrow \angle BOC=\frac{1}{2}\left(\angle A+\angle B+\angle C\right)-\frac{1}{2}\angle A \\ \Rightarrow \angle BOC=\frac{1}{2}\left(180\right)°-\frac{1}{2}\left(40\right)° \\ \Rightarrow \angle BOC=70° \end{gathered}$$