There is incircle $\Gamma$ of triangle $ABC$ tangent to $AB,BC,CA$ respectively at $K,L,M$. Point $D$ is the midpoint of section $MK$. $DL$ is diameter of another circle which intersects with $\Gamma$ at $L,P$ and with $MK$ at $D,R$. Point of intersection line $PR$ with $AD$ is $S$ show that $DS=AS$ .
So I have a huge problem with the adoption of the concept, what I noticed so far is $\angle PLD = \angle DRS$ , $\angle RDS = 90 = \angle LPD $ hence $\triangle LPD \sim \triangle DRS$
