# Given a scalene triangle $ABC$ with $H$ orthocenter, prove that two lines are parallel

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 $$\angle HPQ=\angle HQP=90-\frac {\angle A}{2}$$.