anonymous1173 wrote: $ABC$ be a triangle. Its incircle touches the sides $CB, AC, AB$ respectively at $N_{A},N_{B},N_{C}$. The orthic triangle of $ABC$ is $H_{A}H_{B}H_{C}$ with $H_{A}, H_{B}, H_{C}$ are respectively on $BC, AC, AB$. The incenter of $AH_{C}H_{B}$ is $I_{A}$; $I_{B}$ and $I_{C}$ were defined similarly. Prove that the hexagon $I_{A}N_{B}I_{C}N_{A}I_{B}N_{C}$ is regular. In general, it is not regular. It is just symmetric with respect to a point. Darij

my mistake so sorry.. fixed

Here is my solution. I think it's very nice. First, denote $I$ the incenter of $\triangle ABC$. Second, let $D_{A}$ be the foot of the bisector of $\angle A$. We define $D_{B}$ and $D_{C}$ similarly. Finaly, let $r$ be the radius of the incircle of $\triangle ABC$. Lemma 1. Point symmetric to $H_{C}$ with respect to line $N_{B}N_{C}$ lies on line $CI$. Similarly for $H_{A}$ and $H_{B}$. Proof. Let $P$ be the projection of $B$ onto $CI$ and let $BP$ intersect $AC$ in $Q$. Obviously, $\angle AQB=180^{\circ}-\frac{\angle A}{2}-\frac{\angle B}{2}=\angle AIB.$ So $I$ lies on circumcircle of $\triangle ABQ$, hence line $N_{B}N_{C}$ is Simson line of $\triangle ABQ$ and point $I$, therefore $P$ lies on $N_{B}N_{C}$. As $\angle BH_{C}C=\angle BPC=90^{\circ}$ quadrilateral $BCPH_{C}$ is inscribed. So $\angle N_{C}PH_{C}=90^{\circ}-\angle D_{C}PN_{C}-\angle H_{C}PB=$ $=90^{\circ}-\frac{\angle B}{2}-\angle H_{C}CB=\frac{\angle B}{2}=\angle D_{C}PN_{C}.$ Hence, line $PH_{C}$ is symmetric to line $CI$ with respect to $N_{B}N_{C}$. Thus the Lemma is proved. Solution. Let $\triangle A'H'_{B}H'_{C}$ be symmetric to $\triangle AH_{B}H_{C}$ with respect to line $N_{B}N_{C}$. Obviously point $A'$ lies on line $AI$. From Lemma 1. $H_{B}$ lies on $BI$ and $H_{C}$ lies on $CI$. As $\triangle AH_{B}H_{C}\sim \triangle ABC$ we have $\triangle A'H'_{B}H'_{C}\sim \triangle ABC.$ Moreover, it is easy to verify that $A'H'_{B}\parallel AB,$ $H'_{B}H'_{C}\parallel BC,$ $H'_{C}A\parallel CA.$ Hence, There is a homothety with center $I$ that transforms $\triangle ABC$ into $\triangle A'H'_{B}H'_{C}$. This homothety transforms incenter of $\triangle ABC$ into incenter of $\triangle A'H'_{B}H'_{C}$. But $I$ is transformed into itself. So $I$ is the incenter of $\triangle A'H'_{B}H'_{C}$. Therefore, $I$ is symmetric to $I_{A}$ with respect to line $N_{B}N_{C}$. So $I_{A}N_{B}=I_{A}N_{C}=r.$ We prove similarly for the remaining segments. Thus I have proved that the hexagon $I_{A}N_{B}I_{C}N_{A}I_{B}N_{C}$ has all sides equal to $r$.

Why is $A'H_{B}'$ parallel to $AB$?

nxxx wrote: Why is $A'H_{B}'$ parallel to $AB$? Dear nxxx, line symmetric to $AC$ with respect to $N_{B}N_{C}$ is parallel to AB, because $\angle AN_{B}N_{C}=\angle AN_{C}N_{B}$.

derezins wrote: nxxx wrote: Why is $A'H_{B}'$ parallel to $AB$? Dear nxxx, line symmetric to $AC$ with respect to $N_{B}N_{C}$ is parallel to AB, because $\angle AN_{B}N_{C}=\angle AN_{C}N_{B}$. Thank you very much! Now I understand this perfectly well!

The triangles $\triangle AH_{B}H_{C}\sim \triangle ABC$ are oppositely similar with similarity coefficient $\cos A$ and the bisector $AI \equiv AI_{A}$ of $\angle A$ is common. Let the incircle $(I_{A})$ of the $\triangle AH_{B}H_{C}$ touch $AH_{C}\equiv AB$ at $T_{B}.$ Then $I_{A}T_{B}= IN_{C}\cos A = r \cos A,$ $AT_{B}= (s-a) \cos A = AN_{C}\cos A,$ so that the other leg $T_{B}N_{C}$ of the right triangle $\triangle I_{A}T_{B}N_{C}$ is $T_{B}N_{C}= AN_{C}-AT_{B}= (s-a) (1-\cos A) = \frac{2 r \sin^{2}\frac{A}{2}}{\tan \frac{A}{2}}= r \sin A.$ The hypotenuse is then $I_{A}N_{C}= \sqrt{I_{A}T_{B}^{2}+T_{B}N_{C}^{2}}= r \sqrt{\cos^{2}A+\sin^{2}A}= r$ and similarly, $I_{A}N_{B}= I_{B}N_{A}= I_{B}N_{C}= I_{C}N_{A}= I_{C}N_{B}= r.$