Before I start, I want to say that I already have calculated the correct result of this exercise (on my own) and that I am only interested in finding some formal underpinnings of my calculations.
First we have a circle with some radius r and a center / middle point M, and a triangle with the vertexes A, B and C. The corresponding angles are called a, b and c with a = 44° (belongs to A).
The triangle (A, B, C) is inside of the circle with A and M sharing the same coordinates. Or in other words, the vertex A and the middle point M are the same point with different names. The two remaining vertexes B and C are both on the line of the circle. So the distances between the circle's center M and B, M and C equals the radius r: MB = r, MC = r.
Given this we know that AB = r, and AC = r. We also know that AB = AC which results in our triangle having two sides of the same length. If a triangle has two sides of the same length it is a isosceles triangle.
In an isosceles triangle, two angles are equal. The base is formed by BC, with AB and AC being the legs. The vertex angle is a, and the two base angles are b and c. b and c have to be equal (b = c).
Given that a triangle has a total angle sum of 180°, we know that a = 44° and that b = c, we can calculate b and c as follows:
b + c = 180° - a = 180° - 44° = 136°
b = c = 136° / 2 = 68°
At the end of these calculations we know that a = 44°, b = 68° and c = 68°.
Knowing b and c, I am able to calculate the angles of the two triangles that are inside of the triangle A, B, C. But those calculations are unimportant for my question, therefore I won't bother to describe them.
My real question is, are there official theorems, formulas for calculations like this to underpin my results. Like if there is a triangle (X, Y, Z) with X = M (the center of a circle) and Y and Z are on the line of the circle (XY = XZ = r (radius)) then the triangle is an isosceles triangle?