Let points $A, B$ and $C$ be on a circle $O$ such that segment $AB$ is a diameter of $O$ (meaning the center of the circle is the midpoint of the segment). Prove $m∠C = \frac{1}{2}$ (Internal angle sum(Δ$ABC$)).
Now this is not in Euclidean geometry and it is in neutral geometry. I don't know how to prove it and plus I'm limited since I haven't proven that the sum of the angles in a triangle add up to $180$ and the exterior angle inequality where the angle outside a triangle adds up to the two interior angles. Instead all I have is the Sacceri Legendre theorem (all angles of a triangle add up less than $180^{\circ}$ and exterior angles (angle on the outside is greater than either interior angles). I can use everything else.
Here is a picture:
Draw the line from the center of the circle to $C$. The two triangles are isoceles, so $m \angle C = m \angle A + m \angle B$