Question

In the adjoining figure, the sides $AB$ and $AC$ of $\Delta ABC$ are produced to $D$ and $E$ respectively. If bisector of $\angle CBD$ and $\angle BCE$ meet at $O$, then show that $\angle BOC={90}^o-\frac{\angle A}{2}$.

Answer 1

We have from the figure we have, $\angle OBC+\angle OCB+\angle O={180}^o$........(1).

Also given, $\angle OBC=\frac{1}{2}\angle DBC={90}^o-\frac{1}{2}\angle ABC$.......(2) and $\angle OCB=\frac{1}{2}\angle BCE={90}^o-\frac{1}{2}\angle ACB$.....(3). [ Since $\angle DBC={180}^o-\angle ABC$ and $\angle ECB={180}^o-\angle ACB$]

Now adding (2) and (3) we get,

$\angle DBC+\angle ECB={180}^o-\frac{1}{2}(\angle ABC+\angle ACB)$

or, ${180}^o-\angle BOC={180}^o-\frac{1}{2}({180}^o-\angle BAC)$ [ Using (1)]

or, $\angle BOC={90}^o-\frac{1}{2}\angle BAC$.