A problem involving two orthocenters of two triangles inscribed the same circle

Question

I found out this geometry problem via Geogebra, however, I could not prove it. Please help me.

Thanks.

Let $ABC$ be an acute triangle with $AB < AC$, inscribed $(O)$. Let $T$ be an arbitrary point on the small arc $AB$. Let $H, K$ be the orthocenters of $\Delta ABC$, $\Delta TBC$ respectively. Prove that $ATKH$ is a parallelogram.

Here is the figure

Answer 1

First, $AH$ and $TK$ are parallel because they are both perpendicular to $BC.$

Now let us prove that $AH$ and $TK$ are equal. Let $M$ be the centroid of the triangle $ABC$, $O$ be the center of the circumcircle of the triangle $ABC,$ $E$ be the foot of the perpendicular drawn from $O$ to $BC,$ or the middle of $BC$, in other words. It can be seen that $AH$ is homothetic to $OE$ with M as the center of the homothety and coefficient $-2$. But the very same reasoning can be applied to the triangle $TCB$, since it has the same side $BC$ and the same circumcircle. So $TK$ is also equal to $2\cdot OE$.

We got that $AH$ and $TK$ are equal and parallel, so $AHKT$ is a parallelogram.

Answer 2

We use this property that a circle passing the orthocenter and two vertexes of the triangle has equal diameter to that of circumcircle.As can be seen in the picture we have:

$\overset{\large\frown}{BHC}=\overset{\large\frown}{BFC}\Rightarrow \angle BAC=\angle BKC$

Also:

$\overset{\large\frown}{TB_O}=\overset{\large\frown}{TB_M}\Rightarrow \angle TAB=\angle TKB$

where $_O$ means arc belonging to the circle center at O and M represents the arc on the circle on LHS of the picture. In this way we have:

$\angle TAC=\angle TKC$

AF is perpendicular to BC so we have:

$\overset{\large\frown}{HC}=\overset{\large\frown}{FC}\Rightarrow \angle HAC=\angle HKC\Rightarrow \angle TAH=\angle TKH$

which means ATKH is a parallelogram.

Answer 3

Idea: Show $TA=KH$.

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Let $H',K',O',\odot(O')$ be the reflection of the points $H,K,O$ and of the circumcircle $\odot(O)$ w.r.t. the line $BC$.

From $\Delta BHC=\Delta BH'C$, we obtain the angle in $H'$ in the last triangle, $\hat H'=\hat H=180^\circ - \hat A$, so $H'$ is on the circumcircle $\odot(O)$. The same argument gives $H',K'\in\odot(O)$, $H,K\in\odot(O')$.

Because of $TKK'\|AHH'$ (directions perpendicular on $BC$), $TAH'K'$ is a trapez inscribed in a circle. We conclude $$ TA=K'H'=KH\ , $$ and then immediately $TAHK$ parallelogram. (We have already $TK\|AH$, which should be enough. But if not, chase angles $\widehat{K'TA}=\widehat{TK'H'}=\widehat{KK'H'}=\widehat{K'KH}$ by reflection, so also $TA\|KH$.)

$\square$