Given a scalene triangle $\triangle ABC$ with $H$ the orthocenter of the triangle. The internal bisector of the angle $\angle BAC$ intersects the lines $BH$ and $CH$ at the points $Λ$ and $Θ$ correspondingly. If (c) is the circumcircle of the triangle $\triangle ΗΛΘ$ and (e) is the tangent of (c) at $Λ$. The perpendicular line from $H$ towards (e) is the line (e) at the point $N$. If $T$ is the point where the perpendicular line from $Λ$ towards $ΗΘ$ intersects the line $ΗΘ$, prove that $NT$ and $ΛΘ$ are parallel.
My thoughts are the following:
$\angle ΛΤΗ+\angle ΗΝΛ=90^{o}$
Hence $ΤΗΝΛ$ is inscribable.
Hence we have that $\angle ΤΛΝ+ \angle ΤΗΝ=180^{o}$
And also $\angle ΛΝΤ=\angle ΛΗΤ$
Hence it is enough to prove that $\angle ΛΗΘ = \angle ΛΘΗ$.
Moreover it seems like $\triangle ΛΘΗ$ is equilateral, but I can't prove that. Could you please explain to me how to solve this question?
I will call the points where the angle bisector of $\angle BAC$ intersects $BH$ and $CH$ as $P$ and $Q$ respectively. Let $S$ be some arbitrary point on the right[Since $P$, $Q$ are on the right of $H$ ] of point $P$ on the tangent to the circumcircle of $\triangle HPQ$.
Observe that, $\angle SPQ=\angle PHQ=\angle PHT=\angle PNT=\angle SNT$ and thereafter $PQ\parallel NT$.
$\triangle HPQ$ is not necessarily equilateral but isosceles since $\angle HPQ=\angle HQP=90-\frac {\angle A}{2}$.