In triangle ABC, $\angle A = 70^\circ$. The bisectors of the exterior angles at B and C meet at J. What is $\angle BJC$?

Explanation

1. 01

   The exterior angle at B is $180^\circ - \angle B$ and at C is $180^\circ - \angle C$. Their halves are $90^\circ - \tfrac{1}{2}\angle B$ and $90^\circ - \tfrac{1}{2}\angle C$.

2. 02

   In triangle BJC, $\angle BJC = 180^\circ - (90^\circ - \tfrac{1}{2}\angle B) - (90^\circ - \tfrac{1}{2}\angle C) = \tfrac{1}{2}(\angle B + \angle C)$.

3. 03

   $\angle B + \angle C = 110^\circ$, so $\angle BJC = 55^\circ$.

4. 04

   Shortcut: $\angle BJC = 90^\circ - \frac{1}{2}\angle A = 90^\circ - 35^\circ = 55^\circ$.