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?

  • $\begingroup$ No they aren't. $\endgroup$ – 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.

  • $\begingroup$ What is $\angle ALLB$? (why not just $\angle ALB$?) $\endgroup$ – Pero Dec 3 '18 at 23:16
  • $\begingroup$ @Pero This is a typo. It is not an angle, it was suppose to be a fraction. $\endgroup$ – Futurologist Dec 3 '18 at 23:57
  • $\begingroup$ 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 $CKDL$) $\endgroup$ – Pero Dec 4 '18 at 0:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.