On side \( AC \) of triangle \( ABC \), a point \( P \) is chosen such that \( AP = \frac{1}{3} AC \), and on segment \( BP \), a point \( S \) is chosen such that \( CS \perp BP \). A point \( T \) is such that \( BCST \) is a parallelogram. Prove that \( AB = AT \). Proposed by Bohdan Zheliabovskyi