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, $\angle RPS$ and $\angle RQS$ are equal and $\angle RCS$ is equal to twice the angle at the circumference.

To generalise this theorem to one about $\angle RCS$ 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 $\angle RQS$, because thay lie on the same arc.

$\angle PRQ$ is equal to $\angle PSQ$, because thay lie on the same arc.

$$\alpha + \beta + \gamma =180^o$$ because they are angles of a triangle. $$\gamma + x = 180^o$$ because their arms $PC$ and $CS$ lie on a straight line.

Hence, subtracting the last two equations we obtain $x = \alpha + \beta$, 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.