Let \( BE \) and \( CF \) be the medians of \( \triangle ABC \), and \( G \) be their intersection point. On segments \( GF \) and \( GE \), points \( K \) and \( L \), respectively, are chosen such that \( BK = CL = AG \). Prove that \[ \angle BKF + \angle CLE = \angle BGC. \] Proposed by Vadym Solomka