Question

Suppose that the lines which bisect the exterior angles at $B$ and $C$ of $\Delta ABC$ meet at $D$. Then find $\angle BDC$.

Answer 1

$BD$ is the bisector of $\angle XBC$
$\angle DBC=\frac{1}{2}({180}^o-\angle B)={90}^o-\frac{\angle B}{2}$
Similarly $\angle DB={90}^o-\frac{\angle C}{2}$
In $\Delta BDC$, we have

$\angle BDC={180}^o-\angle DBC-\angle DCB$
$={180}^o-{90}^o+\frac{\angle B}{2}-{90}^o+\frac{\angle C}{2}$
$=\frac{1}{2}(\angle B+\angle C)=\frac{1}{2}({180}^o-\angle A)$
$={90}^o-\frac{\angle A}{2}$