Similar triangles?

Call the intersection of the interior angle bisectors $I$ and the intersection of the exterior angle bisectors $E.$ Let $X$ be an arbitrary point on $AB$ to the right of $B.$

Since $IB$ is the angle bisector of $\angle ABC,$ $\angle IBC = \frac{1}{2} \angle ABC.$

Since $IE$ is the angle bisector of $\angle CBX,$ $\angle CBE = \frac{1}{2} \angle CBX = \frac{1}{2}(180 - \angle ABC) = 90 - \frac{1}{2} \angle ABC.$

Thus, $\angle IBE = \angle IBC + \angle CBE = \frac{1}{2} \angle ABC + \left ( 90 - \frac{1}{2} \angle ABC \right ) = 90^{\circ}.$ A similar process also shows that $\angle ICE = 90^{\circ}.$ This narrows our choices down to either the first or the last one.

Now, we shall find $\angle BIC.$ Since $\angle IBC = \frac{1}{2} \angle B$ and $\angle ICB = \frac{1}{2} \angle C,$ we have

$\begin{aligned} \angle BIC &= 180 - (\angle IBC + \angle ICB) \\ &= 180 - \dfrac{1}{2}(\angle B + \angle C) \\ &= 180 - \dfrac{1}{2}(180 - \angle A) \\ &= 90 + \dfrac{1}{2} \angle A \\ &= 90 + \dfrac{1}{2} (20) \\ &= 100^{\circ}. \end{aligned}$

Thus, we conclude that $BICE$ has angles $\boxed{100^{\circ}, 90^{\circ}, 90^{\circ}, 80^{\circ}}.$