Right triangle $\Delta ABC$ ($\angle ACB=90°$). The following is constructed: from point $C$ altitude $CD$, angle bisector $CL$ of $\angle ACB$, angle bisector $DK$ of $\angle ADC$, angle bisector $DN$ of $\angle BDC$.
$D, L$ lie on $AB$, $K$ lies on $AC$, $N$ lies on $BC$.
Prove that $C, K, L, D, N$ lie on same circle and prove that $|KN|=|CL|$
I think that I need to do something with quadrilateral $CKDN$. I got that $$|KN|=\sqrt{|CK|^2+|CN|^2}=\sqrt{|DK|^2+|DN|^2}$$ $$\angle CKD+\angle CND=180°$$ I also tried to express sides with the angle bisector theorem, but I don't know how to continue / what I need to solve this. How can I solve this problem?