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?
First, since $CD \perp AB$ and $DK$ and $DN$ are angle bisectors to the right angles $\angle \, ADC$ and $\angle \, BDC$, then $$\angle \, KDN = \angle \, KDC + \angle \, NDC = 45^{\circ} + 45^{\circ} = 90^{\circ}$$ However, $\angle \, KCN = 90^{\circ}$ so the quadrilateral $CKDN$ is inscribed in a circle.
Next, prove that $KL\, || \, CB$ and $NL\, || \, CA$ using the properties of angle bisectors and the similarity between triangles $ABC, ACD$ and $BCD$. Indeed, since $DK$ is a bisector of the angle at vertex $D$ of triangle $\Delta \, ADC$, we apply the theorem that $$\frac{AK}{KC} = \frac{AD}{DC}$$ But triangles $\Delta \, ACD$ is similar to $\Delta \, ABC$ so $$\frac{AD}{DC} = \frac{AC}{CB}$$ so $$\frac{AK}{KC} = \frac{AC}{CB}$$ By the fact that $CL$ is an angle bisector of the angle at vertex $C$ of triangle $\Delta\, ABC$ we have that $$\frac{AC}{CB} = \frac{AL}{LB}$$ so consequently $$\frac{AK}{KC} = \frac{AL}{LB}$$ which by Thales' intercept theorem implies that $KL \, || \, CB$. Analogously, one can show that $NL \, || \, CA$.
Then quad $CKLN$ is a rectangle, so $KL =NL$ as diagonals in a rectangle. Therefore the point $L$ also lies on the circumcircle of quad $CKDN$ and $KN$ and $CL$ are diameters of the said circle.