MathNet 2020 — Math Olympiad Vault — Part 06 Question 50

Point $P$ is chosen on the smaller arc $BC$ of the circumscribed circle of an acute-angled triangle $ABC$. Points $R$ and $S$ on sides $AB$ and $AC$ respectively are chosen such that $CPRS$ is a parallelogram. Point $T$ on arc $AC$ of the circumscribed circle of $\triangle ABC$ is such that $BT \parallel CP$. Prove that $\angle TSC = \angle BAC$.

*(Anton Tryhub)*

Editorial note: the source solution refers to Figure 27, but the supplied record contains no corresponding image. The point construction is specified in the text.