Right triangle geometry problem

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?

• No they aren't. – Pero Dec 3 '18 at 12:54

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.
• What is $\angle ALLB$? (why not just $\angle ALB$?) – Pero Dec 3 '18 at 23:16
• Oh I see. I think you also have a typo at the beginning (3rd row). I think it should be $CKDN$ is inscribed in circle (not $CKDL$) – Pero Dec 4 '18 at 0:04