We get $TA$ parallel to $KR$ because $\angle{ATS}=\angle{SJK}=\angle{SRK}$ We extend $KS$ to $P$ where $KS$ intersects $TA$ at $P$ Now, It's easy to prove that $TPRK$ is a rombus So, $\angle{TPK}=\angle{PKR}$ Again, $\angle{ARS}=\angle{SKR}$ So, $\angle{TPK}=\angle{ARS}$ So, $APSR$ is cyclic. $\angl...