Post Number:#2 by Ananya Promi » Sat Jul 29, 2017 12:38 pm
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.
$\angle{PRT}=\angle{RTK}$
$\angle{STK}=\angle{SAT}$
So, $KT$ is tangent to the circle.
We are done