Question

In fig. the sides$AB$ and $AC$ of $\Delta ABC$ are produced to points $E$ and $D$ respectively. If bisectors $BO$ and $CO$ of $\angle CBEand\angle BCD$ respectively meet at point $O$, then prove that $\angle BOC={90}^o-\frac{1}{2}\angle BAC$.

Answer 1

Here $BO,CO$ are the angle bisectors of $\angle EBC$ & $\angle DCB$ intersect each other at $O$.

$\therefore \angle EBO=\angle OBC$ and $\angle OCB=\angle OCD$

Side $AB$ and $AC$ of $△ABC$ are produced to $E$ and $D$ respectively.

$\therefore \angle EBC=\angle BAC+\angle ACB$ --------------(1)

And $\angle DCB=\angle BAC+\angle ABC$ --------------(2)

Adding (1) and (2) we get

$\angle EBC+\angle DCB=2\angle BAC+\angle ABC+\angle ACB$

$2\angle OBC+2\angle OCB=\angle BAC+{180}^{\circ }$ ------Sum of interior angles of a triangle is equal to ${180}^{\circ }$

$\angle OBC+\angle OCB=\frac{1}{2}\angle A+{90}^{\circ }$ ----------(3)

But in a $△BOC=\angle OBC+\angle OCB+\angle BOC={180}^{\circ }$ --------(4)

From equations (3) and (4) we get

$\frac{1}{2}\angle BAC+{90}^{\circ }+\angle BOC={180}^{\circ }$

$\angle BOC={90}^{\circ }-\frac{1}{2}\angle BAC$