This is a very simple generalisation of the well known theorem that the angle at the centre of a circle is twice the angle at the circumference subtended on the same arc. First, as is usually done, take C at the centre of the circle, fix the arc RS and take P and Q to be any points on the circumference. Then the triangles CPR and CQS must be congruent and isosceles because CP, CQ, CR and CS are radii. Hence, the angles RPS and RQS are equal and the angle at the centre RCS is equal to twice the angle at the circumference.

To generalise this theorem to one about the angle RCS where C is any point inside the circle, the triangles are no longer isosceles, nor are they congruent to each other any more.

Aleksander Twarowski from Gdynia Bilingual High School No 3, Poland proves that
angle RCS = angle RPC + angle CRP = angle SQC + angle CQS.

Angle RPS is equal to RQS, because thay lie on the same arc.
Angle PRQ is equal to PSQ, because thay lie on the same arc.

α + β + γ = 180^{o}

because they are angles of a triangle.

γ + x = 180^{o}

because their arms PC and CS lie on a straight line.

Hence, subtracting the last two equations we obtain $x = α + β$ , which is my generalisation of the theorem about the angle at the centre of a circle being twice the angle at the circumference subtended on the same arc.