Where two external bisectors meet: $90^{\circ }-\frac{1}{2}\angle A$

How to use it

1. 1Only $\angle A$ is needed; the individual base angles never enter, so the result holds for every triangle with that apex angle.
2. 2To rebuild it rather than recall it: each external bisector makes $90^{\circ }-\frac{1}{2}\angle B$ (respectively $90^{\circ }-\frac{1}{2}\angle C$) with $BC$, and closing the angle sum in triangle $DBC$ leaves $\frac{\angle B+\angle C}{2}$.
3. 3Pair it with the internal result: the incentre gives $90^{\circ }+\frac{1}{2}\angle A$ and the excentre $90^{\circ }-\frac{1}{2}\angle A$, so the two are supplementary.

Step-by-step

1. 1
   $\angle B+\angle C=180^{\circ }-40^{\circ }=140^{\circ }$

   Why: The angle sum of the triangle.

2. 2
   $\angle BDC=\frac{\angle B+\angle C}{2}=70^{\circ }=90^{\circ }-\frac{40^{\circ }}{2}$

   Why: The two $90^{\circ }$ from the external bisectors cancel against the $180^{\circ }$ angle sum in triangle $DBC$.