Problem

Source: Iranian TST

Tags: TST, Iranian TST, geometry, Iran



In a triangle $ABC$, $\angle A$ is $60^\circ$. On sides $AB$ and $AC$ we make two equilateral triangles (outside the triangle $ABC$) $ABK$ and $ACL$. $CK$ and $AB$ intersect at $S$ , $AC$ and $BL$ intersect at $R$ , $BL$ and $CK$ intersect at $T$. Prove the radical centre of circumcircle of triangles $BSK, CLR$ and $BTC$ is on the median of vertex $A$ in triangle $ABC$. Proposed by Ali Zamani