I divided the solution into two parts: the demonstration of the congruence of angles, and the demonstration that the three angles add up to 180^o.

1. Congruence of angles in a lune

Let O₁ and O₂ be the centres of two circles, A and B the intersection points and M and N the intersections of the tangents to these two circles. I observed that triangles O₁AO₂ and O₁BO₂ are congruent, because they have a common side O₁O₂, and the other two sides are radii in the two circles. So, angle O₁AO₂ and O₁BO₂ are equal to say q. A tangent to a circle and the radius to the point of contact are perpendicular so ÐMAO₂ = ÐNAO₁ = ÐMBO₂ = ÐNBO₁ = 90^o. Hence ÐO₁AM = ÐO₂AN = Ð O₁BM = Ð = ÐO₂BAN = q- 90^o. So, ÐMAN = ÐMBN.

2. Let O₁, O₂ and O₃ be the centres of the three circles respectively, D the point common to all three circles, and T₁T₁ ' , T₂T₂ ' , T₃T₃ ' the three tangents. In my figure the radius and the tangent to circle 1 are in red, to circle 2 in blue and to circle 3 in green respectively.

From the previous point, I see that angle a has the same measure with angle T₂DT₁ ' and to the angle T₁DT₂ ' (the last two being vertically opposite angles).

In the same manner I found: b = ÐT₁ ' DT₃ ' = Ð T₁DT₃ and g = ÐT₃DT₂ = ÐT₃ ' DT2 ' .

This way, it is easy to see that the three angles add up to 180^o: a+ b+ g = T₂DT₁ ' + T₁ ' DT₃ ' + T₃ ' DT₂ ' = T₂DT₂ ' = 180^o.

For the second case, I follow exactly the same steps.

I find a = ÐT₂DT₁ ' , b = ÐT₂ ' DT₃ ' = ÐT₂DT₃ and gÐT₃DT₁ = ÐT₃ ' DT₁ ' .

The result is the same, i.e. that in this case the three angles add up to 180^o again.