Question

In ∆ABC, sides AB and AC are produced to D and E, respectively. BO and CO are the bisectors of ∠CBD and ∠BCE, respectively. Prove that $\angle BOC=90°-\frac{1}{2}\angle A.$

Answer 1

Let
$$\begin{gathered} \angle OBC=\angle OBD=\angle 1\text{and} \\ \angle OCB=\angle OCE=\angle 2 \end{gathered}$$
We have:
$$\begin{gathered} \angle ABC+\angle CBD=180°\left[\text{Linear pair}\right] \\ \Rightarrow \frac{1}{2}\angle ABC+\frac{1}{2}\angle CBD=90° \\ \Rightarrow \frac{1}{2}\angle ABC+\angle 1=90°\left[\because \frac{1}{2}\angle CBD=\angle OBC=\angle OBD\right] \\ \Rightarrow \angle 1=90°-\frac{1}{2}\angle ABC...\left(i\right) \end{gathered}$$

Also,
$$\begin{gathered} \angle ACB+\angle BCE=180°\left[\text{Linear pair}\right] \\ \Rightarrow \frac{1}{2}\angle ACB+\frac{1}{2}\angle BCE=90°\left[\because \frac{1}{2}\angle BCE=\angle OCB=\angle OCE\right] \\ \Rightarrow \frac{1}{2}\angle ACB+\angle 2=90° \\ \Rightarrow \angle 2=90°-\frac{1}{2}\angle ACB...\left(ii\right) \end{gathered}$$
Now, in $∆OBC$, we have:
$$\begin{gathered} \angle OBC+\angle OCB+\angle BOC=180° \\ \Rightarrow \angle 1+\angle 2+\angle BOC=180° \\ \Rightarrow 90°-\frac{1}{2}\angle ABC+90°-\frac{1}{2}\angle ACB+\angle BOC=180°\left[\text{Using}\left(i\right)\text{and}\left(ii\right)\right] \end{gathered}$$

$$\begin{gathered} \Rightarrow \angle BOC=\frac{1}{2}\left(\angle ABC+\angle ACB\right) \\ \Rightarrow \angle BOC=\frac{1}{2}\left(\angle A+\angle ABC+\angle ACB\right)-\frac{1}{2}\angle A \\ \Rightarrow \angle BOC=\frac{1}{2}\left(180\right)°-\frac{1}{2}\angle A\left[\because \mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180°\right] \\ \Rightarrow \angle BOC=90°-\frac{1}{2}\angle A \end{gathered}$$