Question

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

Answer 1

$\angle CBE+y^o={180}^o$ .........(a) {Linear Pair}

$\angle BCD+z^o={180}^o$ .........(b) {Linear Pair}

Ray BO is the bisector of $\angle CBE$.

Therefore, $\angle CBO=\frac{1}{2}\angle CBE$

$=\frac{1}{2}({180}^o-y^o)$ {from (a)}

$={90}^o-\frac{y^o}{2}$ ...... (1)

Similarly, ray CO is the bisector of $\angle BCD$.

Therefore, $\angle BCO=\frac{1}{2}\angle BCD$

$=\frac{1}{2}({180}^o-z^o)$ {from (b)}

$={90}^o-\frac{z^o}{2}$ ...... (2)

In $△BOC,$

$\angle BOC+\angle BCO+\angle CBO={180}^o$ ...... (3)

Substituting (1) and (2) in (3), we get,

$\angle BOC+{90}^o-\frac{z^o}{2}+{90}^o-\frac{y^o}{2}={180}^o$

So, $\angle BOC=\frac{z^o}{2}+\frac{y^o}{2}$

or, $\angle BOC=\frac{1}{2}(y^o+z^o)$ ...... (4)

But, $x^o+y^o+z^o={180}^o$ (Angle sum property of a triangle)

Therefore, $y^o+z^o={180}^o-x^o$

Therefore, (4) becomes

$\angle BOC=\frac{1}{2}({180}^o-x{)}^o$

$={90}^o-\frac{x}{2}$

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