Question

The bisector of $\angle B$ and $\angle C$of triangle $ABC$ intersect each other at the point$O$. Prove that$\angle BOC=90°+\frac{1}{2}\angle A.$

Answer 1

Given, The bisector of $\angle B$ and $\angle C$of triangle $ABC$ intersect each other at the point$O$.

To prove:$\angle BOC=90°+\frac{1}{2}\angle A.$

Proof:

Consider$∆BOC.$in the below figure.

From the angle sum property of a triangle, we have

$\angle 1+\angle 2+\angle BOC=180°......\left(i\right)$

In$∆ABC,$

$\angle A+\angle B+\angle C=180°.....\left(ii\right)$

Since The bisector of $\angle B$ and $\angle C$of a triangle $ABC$ intersect each other at the point$O$

So, $\angle B=2\angle 1$ and $\angle C=2\angle 2$

Substitute the value of $\angle B$ and $\angle C$in (ii), we get

$\angle A+2\angle 1+2\angle 2=180°$

Dividing both sides by $2$we get,

$$\begin{gathered} \frac{1}{2}\angle A+\angle 1+\angle 2=90° \\ \Rightarrow \angle 1+\angle 2=90°-\frac{1}{2}\angle A \end{gathered}$$

Put the above equation in (i), we get

$$\begin{gathered} 90°-\frac{1}{2}\angle A+\angle BOC=180° \\ \Rightarrow \angle BOC-\frac{1}{2}\angle A=180°-90° \\ \Rightarrow \angle BOC=\frac{1}{2}\angle A+90° \end{gathered}$$

Thus, $\angle BOC=90°+\frac{1}{2}\angle A.$

Hence proved.